Compound Nomenclature Practice

Compound Nomenclature Practice boron trichloride CCl4 N2O3 Silver chromate HClO2 Ferric oxide Sodium chlorate Cu(NO3)2 potassium permaganate HNO3 plumbous nitrite Fe(ClO4)3 BCl3 carbon tetrachloride dinitrogen trioxide Ag2(CrO4) chlorous acid Fe2O3 NaClO3 Copper II nitrate KMnO4 Nitric Acid Pb(NO2)2 Iron III perchlorate

Chemists have defined the unified atomic mass unit as the unit of mass for atoms: The amu is defined as: 1 12 C atom = = 12 amu therefore... 1 amu = 1 x Mass of 1 12 C atom 12

Masses written in the periodic table are average 12 isotopic masses and relative to carbon-12 6 C A He atom is 4 times heavier than a H atom A C atom is 3 times heavier than a He atom A F atom is 19/12 times heavier than a C atom

For more than 150 years chemists have measured relative weights of atoms and not absolute masses as atoms are much to small to be weighed directly on a scale.

The relative average isotopic atomic masses are provided for each element in the periodic table.

A chemical formula of a compound indicates the number ratios of combining atoms. Subscripts: indicate the # of bonded atoms Leading coefficients: indicate the number of molecules or atoms.

The Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a single molecule. SO 2 1S 2O SO 2 32.07 amu + 2 x 16.00 amu 64.07 amu The formula mass is the sum of atomic masses (in amu) in a single ionic formula unit. NaCl 1Na 23.00 amu 1Cl 35.45 amu NaCl 58.45 amu

Atoms, molecules and formula units are very very very small particles which cannot be counted or weighed individually or directly. How can we count numbers of atoms or molecules or formula units if we can not see them to count them?

Single molecules are useful images conceptually, but we can t see them, and the mass is to small to be of practical use in a laboratory. We need many many more molecules to measure on human weighing scales + 2H2 + O2 2H2O 4 amu + 32 amu 36 amu How can we connect amu to grams?? grams +? grams? grams

We can count by comparing masses of the same number of marbles gives ratio of masses Masses 12 red marbles 84 g 12 yellow marbles 48 g 12 purple marbles 40.g 12 green marbles 20.g A single red marble must be 1.75 times heavier than a yellow marble (84/48 =1.75)

Scientists have defined a very large counting number called the mole. 1 Mole = # of 12 C atoms in exactly 12 = 6.023 x 10 23 grams of 12 C Avogadro s Number Determined by experiment!

You can think of the mole as just a counting unit for a collection of 6.02 x 10 23 things. 12 H2O molecules 144 H2O molecules Counting Unit 1 dozen H2O molecules 1 gross H2O molecules 12 X 6.02 x 10 23 molecules 1 mole H2O molecules 6.02 x 10 23 eggs

The mole bridges the mass of 1 atom in amu to the mass of 1 mole of atoms in grams. 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole = 6.02 x 10 23 12 C atoms

The mole links Avogadro s Number of atoms or molecules to their mass in grams. A drop of water the size of the period at the end of this sentence would contain 10 trillion water molecules. Instead of talking about trillions and quadrillions of molecules (and more), it's much simpler to use the mole. 1 mole H2O = 18 grams of water. MOLE

Both sides of the balance have 1 mole of atoms: 6.022 x 10 23 atoms each---but they have different masses. 55.85 g Fe 32.07 g S The mole allows us to count out exactly the same number of things but use mass to do it! 55.85 g Fe = 6.022 x 10 23 atoms Fe = 1 mol Fe atoms 32.07 g S = 6.022 x 10 23 atoms S = 1 mol S atoms

1 Mole of a few substances with their corresponding molar mass. CaCO 3 100.09 g All of these have have exactly A.N. of number of elements or molecules but each have a different mass! Oxygen 32.00 g Water 18.02 g Copper 63.55 g

Chemical reactions require specific whole numbers of reactants in order to produce a specific whole number of products. + 2 molecule H2 + 1 molecule O2 2H2O molecules 2H2 + O2 2H2O 4 amu + 32 amu 36 amu How can we connect amu to grams?? grams +? grams? grams

Therefore, the periodic table tells us the mass of 1-atom of any element and it also tells us the mass of 1-mole of any element in grams.

The key riffs that you have to understand and make use of like a pencil. For all elements, the atomic mass (amu) = molar mass (grams) For any molecule, the molecular mass (amu) = molar mass (grams) For any formula unit, the formula mass (amu) = molar mass (grams)

How Big is Avogadro s Number? ---If a computer could count to one billion in one second it would take the same computer 20 million years to count to Avogadro s Number. ---Spreading 1 mole of marbles over the entire surface of the earth would produce a blanket of marbles almost 5 km high.

Molecular and Molar Mass (Covalent) The Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a single molecule. SO 2 1S 2O SO 2 32.07 amu + 2 x 16.00 amu 64.07 amu The molar mass of a compound is the numerically the same of the molecular mass with units of grams per mole. 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2

Formula Mass-Ionic Compounds The formula mass (or forumula weight) is the sum of the atomic masses (in amu) in a formula unit. 1 Na 1 Cl NaCl 23.00 amu 35.45 amu 58.45 amu The molar mass of an ionic compound is numerically equal to the formula mass, in units of grams/mole. Molar mass of NaCl = 58.45 g/mol 1 mole of NaCl(s) = 58.45 grams NaCl(s) 6.022 10 23 NaCl units = 58.45 grams NaCl(s)

Summary of Mass Terms Term Definition Unit Isotopic mass Mass of an isotope of an element amu Atomic mass (also called atomic weight) Molecular or formula mass (also called molecular weight) Molar mass (M) (gram-molecular weight) Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) Mass of 1 mole of chemical entities (as atoms, ions, molecules or formula units) amu amu g/mol

Do You Understand Molecular and Molar Mass? What is the molecular mass of glucose, C 6 H 12 O 6 Molecular formula C 6 H 12 O 6 Empirical formula CH 2 O Molecular Mass: 6 x 12.011 + 12 x 1.007 + 6 x 15.99 = 180.18 amu Molar Mass: 1 mol C6H12O6 = 180.18 grams C6H12O6 1 mol C6H12O6 = 6.022 X 10 23 molecules C6H12O6

Learning Check: Tabulate the available information in the chemical formula of glucose C 6 H 12 O 6 (MW = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) Atoms in 1 molecule 6 atoms 12 atoms 6 atoms Mass of 1 molecule 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu Atoms/mole of compound 6(6.022 x 10 23 ) atoms 12(6.022 x 10 23 ) atoms 6(6.022 x 10 23 ) atoms Moles of atoms in 1 mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Mass/mole of compound 72.06 g 12.10 g 96.00 g

A chemical formula provides lots of information: the elements in the compound, the ratios elements combined, the mass of one formula or molecular unit, and the mass in grams of 1 mole of compound in grams. Al2(SO4)3 2 atoms of Al and 3 molecules of (SO4) 2-2 moles of Al and 3 moles of (SO4) 2-1 formula unit Al2(SO4)3 = 342.17 amu Al2(SO4)3 1 mole Al2(SO4)3 = 342.17 g Al2(SO4)3 1 mole Al2(SO4)3 = 6.022 x 10 23 formula units Al2(SO4)3 1 mole Al2(SO4)3 = 2 mol Al 3+ ions 1 mole Al2(SO4)3 = 3 mol (SO4) 2- ions 1 mole Al2(SO4)3 = 12 mol O atoms = 12 x A.N. O atoms 1 mole Al2(SO4)3 = 3 mol S atoms = 3 x A.N. S atoms 1 mole Al2(SO4)3 = 2 mol Al atoms =2 x A.N. Al atoms

Molecular formulas indicates the type and the ratios of combining atoms in a covalent compound. H2O(l) covalent compound There are many conversion factors in a formula. 1 molecule H2O = 1 atom O and 2 atoms of H Molecular mass H2O = (2 x 1.008)+15.99 = 18.00 amu Molar mass = 18.00 g/mol H2O 1 mole H2O = 6.022 x 10 23 molecules H2O 1 mole H2O = 2 mol H atoms = 2 x 6.02 x 10 23 H atoms 1 mole H2O = 1 mol O atoms = 6.02 x 10 23 O atoms

We must learn how to recognize conversion factors within a chemical formula, and how to convert grams to moles to molecules to atoms and vis versa. MASS(g) of compound Molar mass (g/mol) MOLES of compound MOLECULES or FORMULA UNITS Chemical Formula Avogadro s Number MOLES of elements in compound ATOMS in a compound

What s In A Chemical Formula? Urea, (NH2)2CO, is a nitrogen containing compound used as a fertilizer around the globe? Calculate the number of moles of urea, # of molecules of urea, # hydrogen atoms present in 25.6 g of urea.

Solution To Urea Problem 1. Calculate the molar mass (MM) of urea, (NH2)2CO MM(NH2)2CO = 2 x MN + 4 x MH + MC + MO MM(NH2)2CO = (2 x 14.07g) + 4 (1.007g) + 12.01g + 15.99g MM(NH2)2CO = 60.06 g 2. # moles of (NH2)2CO in 25.6 g Mol (NH 2 ) 2 CO = 25.6 g (NH 2 ) 2 CO 1 mol (NH 2 ) 2 CO 60.06 g (NH 2 ) 2 CO = 0.426 mol (NH 2 ) 2 CO 3. # molecules of (NH2)2CO in 25.6 g Molec urea = 0.426 mol urea 6.02 1023 molecu urea 1 mol urea = 2.57 10 23 molecu urea 4. # H atoms in 25.6 g (NH2)2CO #H atoms = 0.426 mol urea 4 mol Hatoms 1 mol urea 6.02 1023 H atoms 1 mol urea = 1.03 10 24 atoms

Borax is the common name of a mineral sodium tetraborate, an industrial cleaning adjunct, Na2B4O7. Suppose you are given 20.0 g of borax (a) what is the formula mass of Na2B4O7 (b) how many moles of borax is 20.0 g? (b) how many moles of boron are present in 20.0 g Na2B4O7? (c) how many grams of boron are present in 20.0 g Na2B4O7? (d) how many atoms of boron are present? (e) how many grams of atomic oxygen are present?

Borax is the common name of a mineral sodium tetraborate, an industrial cleaning adjunct, Na2B4O7. Suppose you are given 20.0 g of borax Solution: (a) The formula weight of Na2B4O7 is (2 23.0) + (4 10.8) + (7 16.0) = 201.2. b) # mol borax = (20.0 g borax) / (201.2 g mol 1 ) = 0.10 mol of borax, c) # mol B = 20.0 g borax)/(201.2 g mol 1 ) X 4 mol B/ 1 mol borax = 0.40 mol of B. d) # g B = (0.40 mol) (10.8 g B/ mol B) = 4.3 g B e) # atoms B = 0.40 mol B x 6.02 X 10 23 atoms B/1 mol B = 2.41 X 10 23 atoms B f) # g O = 20.0 g borax)/(201.2 g mol 1 ) X 7 mol O/ 1 mol borax x 15.99 g O/1 mol O = 11.1 g O

Empirical and Molecular Formulas Molecular Formula The formula of a compound as it actually exists according to experimental data. It is a multiple of the empirical formula. CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4 C 5 H 10 O 5 30.02 60.05 90.08 120.10 150.13 Empirical Formula The simplest formula for a compound that gives rise to the smallest set of whole numbers of atoms. CH 2 O

Compounds that have the same % mass of its elements have the same empirical formula! All compounds below have the same % by mass: 40.0% C 6.71% H 53.3% O. Name Molecular Formula Whole-Number Multiple M (g/mol) Use or Function formaldehyde acetic acid lactic acid erythrose ribose glucose CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4 C 5 H 10 O 5 C 6 H 12 O 6 1 2 3 4 5 6 30.03 60.05 90.09 120.10 150.13 180.16 disinfectant; biological preservative acetate polymers; vinegar(5% soln) sour milk; forms in exercising muscle part of sugar metabolism component of nucleic acids and B 2 major energy source of the cell

A chemical formula determines the % mass of each element in a compound. n x molar mass of element molar mass of compound x 100% n is the number of moles of an element in 1 mole of the compound. C 2 H 4 O2 2 x (12.01 g) %C = 60.05 g 4 x (1.008 g) %H = 60.05 g 2 x (16.00 g) %O = 60.05 g x 100% = 40.0% x 100% = 6.714% x 100% = 53.28% 40.0% + 6.71% + 53.2% = 100.0%

A laboratory technique called elemental analysis can determine the % by mass of each element in a compound. Mass Percent Empirical Formula %C = a% %H = b% %O = c% %C = 52.14% %H = 13.13% %O = 34.73% Atomic Mass Atomic Mass C x H y Oz ratio of moles C 2 H 6 O

If we know the % mass of elements in a compound we can determine the empirical formula. With molar mass we get molecular formula Molecular formula GIVEN Molar Mass

Determining the Empirical Formula from Masses of Elements Elemental analysis of a sample of an ionic compound gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this compound? SOLUTION: 2.82 g Na x mol Na = 0.123 mol Na 22.99 g Na 4.35 g Cl x mol Cl = 0.123 mol Cl 35.45 g Cl 7.83 g O x mol O 16.00 g O = 0.489 mol O Na 1 Cl 1 O 3.98 NaClO 4 NaClO 4 is sodium perchlorate.

Determining a Molecular Formula from Elemental Analysis and Molar Mass Dibutyl succinate is an insect repellent used against household ants and roaches. Elemental analysis or analysis indicates that % mass of the composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate?

Step 1: Determine the mass of each element. Assume a 100 gram sample and use the given data we have: C 62.58 g H 9.63 g O 27.79 g Step 2: Convert masses to amounts in moles. Step 3: Write a tentative empirical formula. C 5.21 H 9.55 O 1.74 Step 4: Convert to small whole numbers. Divide by smallest number of moles C 2.99 H 5.49 O

Step 5: Convert to a small whole number ratio. Multiply by 2 to get C 5.98 H 10.98 O 2 The empirical formula is C 6 H 11 O 2 Empirical formula mass = 6(12.01) + 1.008(11) + 2 (16.00) = 115 u. Step 6: Now using the empirical formula mass and molecular mass together determine the molecular formula. Empirical formula mass is 115 u. Molecular formula mass is 230 u. n = Molecular mass/empirical mass = 230 amu/115 amu = 2 The molecular formula is C 12 H 22 O 4

Determining a Molecular Formula from Elemental Analysis and Molar Mass During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness at fatigue. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. Understand what is asked: What is the formula CxHyOz Mass Percent Empirical Formula Molecular Mass Molecular Formula

Determining a Molecular Formula from Elemental Analysis and Molar Mass 1. Assume there are 100. g of lactic acid then use % mass: 40.0 g C 1 mol C 6.71 g H 1 mol H 53.3 g O 12.01g C 1.008 g H 1 mol O 16.00 g O 3.33 mol C 6.66 mol H 3.33 mol O 2. The red numbers are the number of moles of atoms in lactic acid. This is what we use in the formula C 3.33 H 6.66 O 3.33 C 3.33 H 6.66 O 3.33 3.33 3.33 3.33 molar mass of lactate mass of CH 2 O CH 2 O 90.08 g 30.03 g empirical formula 3. The molecular formula will be a whole number multiple of the empirical formula determined BY THE MOLAR MASS ratio 3 C 3H 6 O 3 is the molecular formula

Chemical equations are symbolic representations of what happens in a chemical reaction. CH4(g) + O2(g) CO2(g) + H2O(g) Reactants Yields Products Phases specified: (s) for solids, (l) for liquids, (g) for gases, (aq) for aqueous. Conservation of mass = Balanced equations.

We must know how what information is available in a balanced chemical equation! molecules mass of atoms amount (mol) mass (g) C 3 H 8 (g) + 5O 2 (g) 1 molecule C 3 H 8 + 5 molecules O 2 44.09 amu C 3 H 8 + 160.00 amu O 2 1 mol C 3 H 8 + 5 mol O 2 44.09 g C 3 H 8 + 160.00 g O 2 3CO 2 (g) + 4H 2 O(g) 3 molecules CO 2 + 4 molecules H 2 O 132.03 amu CO 2 + 72.06 amu H 2 O 3 mol CO 2 + 4 mol H 2 O 132.03 g CO 2 + 72.06 g H 2 O total mass (g) 204.09 g 204.09 g

Chemical equations are balanced by trial and error! Example: Ethane, C2H6, reacts with O2 to form CO2 and H2O. Write a balanced equation for this reaction. 1. Write the correct formula(s) for reactants and products. C 2 H 6 + O 2 CO 2 + H 2 O 2. Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2

C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left C 2 H 6 + O 2 2 hydrogen on right 2CO 2 + 3H 2 O multiply H 2 O by 3 4. Balance those elements that appear in two or more reactants or products. C 2 H 6 + O 2 2 O on left 4 O 2CO 2 + 3H 2 O C 2 H 6 + 7 O 2 2CO 2 + 3H 2 O 2 2C 2 H 6 + 7O 2 + 3 O = 7 oxygen on right 4CO 2 + 6H 2 O multiply O 2 by 7 2 remove fraction multiply by 2

4. Balance those elements that appear in two or more reactants or products. C 2 H 6 + O 2 2CO 2 + 3H 2 O 2 O on left 4 O + 3 O = 7 oxygen on right C 2 H 6 + 7 O 2 2CO 2 + 3H 2 O 2 multiply O 2 by 7 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O remove fraction multiply both sides by 2 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O

Balancing Equations: Please try it. Na3PO4(aq) + HCl(aq) ==> H3PO4(aq) + NaCl(aq) Ba(OH)2(aq) + HCl(aq) ==> H2O(l) + BaCl2(aq) C7H14 + O2 ===> H2O(l) + CO2(g)

Solutions Na3PO4(aq) + 3HCl(aq) ====> H3PO4(aq) + 3NaCl(aq) Ba(OH)2(aq) + 2HCl(aq) ====> 2H2O(l) + BaCl2(aq) 2C7H14 + 21O2 ====> 14H2O(l) + 14CO2(g)

Quiz 2 Please put your books away, and take out 1/2 sheet of paper

Quiz 2 1. Write the chemical formula of the compound formed between Al and O? 2. What is the name of the following compound, N2O5? 3. How many oxygen atoms are there in 850. mg of colbalt (III) oxide? Use the proper number of significant digits in your answer.

Only balanced chemical equations contain useful stoichiometric conversion factors. NH3 + O2 ===> NO + H2 Not Balanced 1 moles NH3 = 1 moles O2 6NH3 + 3O2 ===> 6NO + 9H2 Balance first! Correct Stoichiometric Conversion Factors 6 mol NH3 = 3 mol O2 6 mol NH3 = 9 mol H2 6 mol NH3 = 6 mol NO 3 mol O2 = 6 mol NO 3 mol O2 = 9 mol H2 6 mol NO = 9 mol H2

Steps to mastering stoichiometry! 1. First, always write a balanced chemical equation. 2. Work with moles---not masses...why? 3. Use correct stoichiometric factors Molar Mass Grams of Reactant Moles of Reactant balanced equation Grams of Product Moles of Product Molar Mass

Iron III oxide reacts with carbon monoxide as shown below. How many CO molecules are required to react with 25 formula units of Fe 2 O 3 as shown below in the balanced equation? Fe2O3 (s) + CO(g) Fe(s) + CO2(g) Translate words to formula Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g) Balance THE MAGIC 1. Balance. 2. What does the question want? 3. Find stoichiometric factors 4. Use the factor-label method and solve 5. Be mindful of significant figures 6. Check answer

How many CO molecules are required to react with 25 formula units of Fe 2 O 3 as shown below in the balanced equation? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g)

Iron and carbon dioxide form by reaction between iron(iii) oxide and carbon monoxide. How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g)

How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g)

What mass of CO is required to react with 146 g of iron (III) oxide? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g)

What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g)

What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? Fe2O3 (s) + 3CO(g) 2Fe(s) + 3CO2(g)

Methanol,CH 3 OH, burns when ignited in air. Assuming excess O2 is added and 209. g of methanol is combusted, what mass of water and CO2 is produced?

Methanol, CH 3 OH, burn when ignited in air. Assuming excess O2 is added and 209 g of methanol is combusted what mass of water and CO2 is produced? 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O grams CH 3 OH moles CH 3 OH moles H 2 O grams H 2 O grams CH 3 OH moles CH 3 OH moles CO2 grams CO2 209 g CH 3 OH 1 mol CH 3 OH x 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH 18.0 g H 2 O x = 1 mol H 2 O = 235 g H 2 O

Another stoichiometry example from Silberberg Copper metal is obtained from copper(i) sulfide containing ores in multistep-extractive process. After grinding the ore into fine rocks, the first step is to heat it strongly with oxygen gas to form powdered cuprous oxide and gaseous sulfur dioxide. (0) Write a balanced chemical equation for this process (a) How many moles of oxygen are required to fully roast 10.0 mol of copper(i) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(i) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(i) oxide?

(0) Write a balanced chemical equation for this process Cu 2 S(s) + O 2 (g) Cu 2 O(s) + SO 2 (g) unbalanced 2Cu 2 S(s) + 3O 2 (g) 2Cu 2 O(s) + 2SO 2 (g) (a) How many moles of oxygen are required to roast 10.0 mol of copper(i) sulfide? 3 mol O mol O 2 2 mol Cu 2 S = 15.0 mol O 2 =? = 10.0 mol Cu 2 S x 2 (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(i) sulfide is roasted? g SO 2 = 10.0 mol Cu 2 S x 2mol SO 2 2mol Cu 2 S x 64.07g SO 2 1 mol SO 2 = 641g SO 2

2Cu 2 S(s) + 3O 2 (g) 2Cu 2 O(s) + 2SO 2 (g) (c) How many kilograms of oxygen are required to form 2.86 kg of copper(i) oxide? kg O2 = 2.86 kg Cu 2 O x 103 g Cu 2 O kg Cu 2 O x 1 mol Cu 2 O x 143.10g Cu 2 O 3mol O 2 x 2mol Cu 2 O kg O 2 = 0.959 = 1 kg O 2 32.00g O x 2 1000 g O 2 1 mol O 2

The limiting reagent is the reactant that is runs out (is consumed) and determines the quantity or amount of product that can be formed. We need to count the number of reactants to figure out what will limit the amount of cars produced--it s not how much they weigh!

We deal with limiting reagents all the time, we just don t give it a buzzword like chemists do. Reactants Product 168/2 = 84 328/4 = 82

Which molecular gas is the limiting reagent?

Do You Understand Limiting Reagent II? If 25.0 g CH 4 is combusted with 40.0 g O 2, which reactant is the limiting reactant? Notice the absence of excess and there are two reactant masses in the problem = limiting reagent! There are two ways you can calculate the answer. Method 1: For both reactants, use balanced equation & implied stoichiometric factors and compute the amount of any product formed. Method 2: Pick one of the reactant and comute how much of the other reactant you need and compare with the given amount.

If 25.0 g CH 4 is combusted with 40.0 g O 2, which reactant is the limiting reactant? CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) Method 1 (Calculating the # moles of product formed by each reactant to determine which reactant makes the least amount) 1. Let s use the amount of CO2 formed as our yardstick of how much product can be made (we could chose H2O). mol CO 2 = 25.0 g CH 4 1 mol CH 4 16.04 g CH 4 1 mol CO 2 1 mol CH 4 = 1.559 mol CO 2 mol CO 2 = 40.0 g O 2 1 mol O 2 32.00 g O 2 1 mol CO 2 2 mol O 2 = 0.625 mol CO 2 O2 must be the limiting reagent as the amount of CO2 produced is the least amount of product!

If 25.0 g CH 4 is combusted with 40.0 g O 2, which reactant is the limiting reactant? CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) Method 1I (Directly comparing amounts of reactants given in the problem to which is the limiting reagent--less steps) g O 2 needed = 25.0 g CH 4 1 mol CH 4 16.04 g CH 4 2 mol O 2 1 mol CH 4 32.00 g O 2 1 mol O 2 = 99.75 g O 2 O2 is the limiting reagent as we need 99.75 g of it but we are are only given 40.0 g O2! Thus, the amount of product that can be formed is determined by the amount of O2 not by the amount of methane, CH4.

Do You Understand Limiting Reagents? 124.0 g of solid Al metal is reacted with 601.0 g of iron(iii) oxide to produce iron metal and aluminum oxide. Calculate the mass of aluminum oxide formed. 1. Write a balanced equation for all problems 2. Two reactant masses and no excess = limiting reagent. 3. Work in moles (grams => moles => equation stoichiometry) 4. Determine maximum theoretical amount of product for both reactants. The limiting reagent is the one that produces the least.

124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe g Al mol Al mol Al2SO3 produced 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al 101.96 g Al 2 O x 3 = 234 g Al 1 mol Al 2 O 2 O 3 3 g Fe 2 O 3 mol Fe 2 O 3 mol Al2SO3 produced 601 g Fe2O3 x 1 mol Fe 2 O 3 160. g Fe 2 O 3 x 1 mol Al 2 O 3 1 mol Fe 2 O 3 101.96 g Al 2 O x 3 = 383 g Al 1 mol Al 2 O 2 O 3 3 Al make the least and is the limiting reagent! 234 g Al 2 O 3 can be produced

Do You Understand Limiting Reagents? Phosphorus trichloride is a commercially important compound used in the manufacture of pesticides. It is made by the direct combination of phosphorus P4 and gaseous chlorine. Suppose 323 g of chlorine is combined with 125 g P4. Determine the amount of phosphorous trichloride that can be produced when these reactants are combined.

P 4 (s) + Cl 2 (g) PCl 3 (l) P 4 (s) + 6Cl 2 (g) 4PCl 3 (l) Step 1. Converts words to formulas by knowing the nomenclature Step 2. Balance the equation Step 3. Determine the reactant that produces the least product. mol PCl 3 = 323 g Cl 2 1 mol Cl 2 70.91 g Cl 2 4 mol PCl 3 6 mol Cl 2 = 3.04mol PCl 3 mol PCl 3 = 125 g P 4 1 mol P 4 123.88 g P 4 4 mol PCl 3 1 mol P 4 = 4.04mol PCl 3 Step 4. Cl2 is the limiting reagent and determines the amount of PCl3 g PCl 3 = 125 g P 4 1 mol P 4 123.88 g P 4 4 mol PCl 3 6 molcl 2 137.32 g PCl 3 1 mol PCl 3 = 417.5 g PCl 3

Try it At Home:Answer is in your book! Hydrazine(N 2 H 4 ) and dinitrogen tetraoxide(n 2 O 4 ), ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 10 2 g of N 2 H 4 and 2.00 x 10 2 g of N 2 O 4 are mixed?

Try it At Home: Hydrazine(N 2 H 4 ) and dinitrogen tetraoxide(n 2 O 4 ), ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 10 2 g of N 2 H 4 and 2.00 x 10 2 g of N 2 O 4 are mixed? BALANCE! 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O(l) 1.00 x 10 2 mol N g N 2 H 2 H 4 4 = 3.12mol N 2 H 4 32.05g N 2 H 4 3 mol N 2 mol N 2 =? = 3.12 mol N 2 H = 4.68 mol N 4 X 2 2 mol N 2 H 4 2.00 x 10 2 1 mol N g N 2 O 2 O 4 4 X 92.02 g N 2 O 4 = 2.17 mol N 2 O 4 N 2 H 4 is the limiting reactant because it produces less product, N 2, than does N 2 O 4. mol N 2 = 2.17 mol N 2 O 4 X 3 mol N 2 1 mol N 2 O 4 = 6.51 mol N 2 4.68mol N 2 mol N 2 28.02g N 2 = 131g N 2

When we do chemical reactions in the classroom they are ideal and theoretically. Real lab have side-reactions and reduce the amount of actual product obtained vs the theoretical amount. A + B (reactants) C (main product) D (side products) % Yield = obtained laboratory data Actual Amount x 100 Theoretical Amount Calculated from limiting reagent

The % Yield of a chemical reaction is the ratio of product mass obtained in the lab over the theoretical (i.e. calculated) X 100. % Yield = Actual Amount Theoretical Amount x 100 Actual Amount is the amount of product obtained from a reaction in the laboratory. It s given in a word problem. Theoretical Amount is the amount of product that would result if all the limiting reagent reacted.

Learning Check: Calculating Percent Yield Silicon carbide (SiC) is made by allowing silicon dioxide to react with powdered carbon (C) under high temperatures. Carbon monoxide is formed as a byproduct. Suppose 100.0 kg of silicon is processed in the lab and 51.4 kg of SiC is recovered What is the percent yield of SiC from this process? Notice how the word excess is missing and the amount of the other reactant. Must assume other reactant is excess!

Learning Check: Calculating Percent Yield SiO 2 (s) + C(s) SiC(s) + CO(g) 1. Converts words to formulas SiO 2 (s) + 3C(s) SiC(s) + 2CO(g) 2. Balance 3. We are given one reactant, we must assume excess of the other (C) mol SiO 2 = 10 2 kg SiO 2 103 g SiO 2 1 kg SiO 2 1 mol SiO 2 60.09 g SiO 2 1 mol SiC 1 mol SiO 2 = 1664 mol SiC kg SiC = 1664 mol SiCl 40.10 g SiC 1 mol SiC 1 kg SiC 1000 g SiC = 66.73 kg SiC = 66.73 kg SiC 4. We get the actual from experiment and theoretical from calculation and plug them into the yield equation (units of mass should be the same) % Yield = kg Actual kg Theoretical 100 = 51.4 kg 66.73 kg 100 = 77.0%

Putting it all together: Limiting/Percent Yield A student reacts 30.0 g benzene, C 6 H 6, with 65.0 g bromine, Br 2, to prepare bromobenzene, C 6 H 5 Br in the lab (MW C6H6 = 78.102, C 6 H 5 Br 156.99, Br2 = 159.808). After the reaction was complete, the student recovered 56.7 g C 6 H 5 Br. Determine the limiting reagent, the theoretical yield, and the overall % yield? 1. Write a balanced equation 2. Use stoichiometry in balanced equation 3. Find g product predicted limiting reagent actual yield/theoretical yield x 100 4. Calculate percent yield

Learning Check: Calculating Percent Yield A student reacts 30.0 g benzene, C 6 H 6, with 65.0 g bromine, Br 2, to prepare bromobenzene, C 6 H 5 Br. (MW C6H6 = 78.102, C 6 H 5 Br 156.99, Br2 = 159.808). After the reaction was complete, the student recovered 56.7 g C 6 H 5 Br. Determine the limiting reagent, the theoretical yield, and the overall % yield? Step 1. Converts words to formulas and balance the equation C 6 H 6 + Br 2 C 6 H 5 Br + HBr Step 2. Determine the reactant that produces the least product. g C 6 H 5 Br = 30.0 g C 6 H 6 1 mol C 6 H 6 78.1 g C 6 H 6 1 mol C 6 H 5 Br 1 molc 6 H 6 156.9 g C 6 H 5 Br 1 mol C 6 H 5 Br = 60.3 g C 6 H 5 Br g C 6 H 5 Br = 65.0 g Br 2 1 mol Br 2 159.8 g Br 2 1 mol C 6 H 5 Br 1 mol Br 2 156.9 g C 6 H 5 Br 1 mol C 6 H 5 Br = 63.8 g C 6 H 5 Br Step 3. C6H6 is the limiting reagent. 60.3 g C6H5Br is theoret yield % Yield = g Actual g Theoretical 100 = 56.7 g 60.3 g 100 = 94.0%

A solution is a homogenous mixture of a solvent plus a solute (focus is aqueous). The solute is the substance(s) dissolved in the solvent (salt in this case). A solvent is the substance present in the larger amount----water typically in aqueous solutions. Solution is the solute + solvent.

Seawater is a has many dissolved solutes in a water solvent. 70+ dissolved components but only 6 make up >99% the solids 35 grams of dissolved salts per kilogram of seawater SO4 2- Mg 2+ Na + Ca 2+ and K + Cl -

Air is a homogeneous solution of gasses. 21% 78%

We can make homogeneous solutions in different ways. Kind of Solution Gas in Gas Gas in Liquid Gas In Solid Liquid in Liquid Liquid in Solid Solutes Air (O2 + N2 + Ar + CO2) Soda pop H2 in Palladium metal Gasoline (mixture of hydrocarbons) Hg metal in silver (amalgam) Solid in Liquid Seawater (Na + + Mg 2+ +...) Solid in Solids Metal alloys (22k gold)

Here the solute is the liquid purple dye and the solvent is water. Properties include: Distribution of solute in solvent is uniform Components do not separate or sediment on standing Not separable by filtration Solute/Solvent mix in ratios up to the solubility limit of solute

Chemists express the concentration of a solution is the amount of solute present in a given quantity of solvent or solution. grams solute molar mass solute connect the dots M = molarity = moles of solute total liters of solution Density = Mass/Volume

We need to be able to feel concentration and see it for problem solving...it s not hard if you make a picture and visualize it. 1. Remember the defintion of molarity and see it. M = molarity = moles of solute total liters of solution grams solute Molar Mass moles solute Volume of solution Molarity

Calculate the molarity of a solution that contains 12.5 g of pure sulfuric acid (H2SO4) in 1.75 L of solution.

Calculating the Molarity of a Solution Sucrose has a molar mass of 342.29 g/mol. What is the molarity of an aqueous solution made by placing 75.5 g of 85% pure sugar in 500.00 ml of water?

Calculating the Molarity of a Solution Sucrose has a molar mass of 342.29 g/mol. It is a fine, white, odorless crystalline powder with a pleasing, sweet taste. What is the molarity of an aqueous solution made by placing 75.5 g of 85% pure sugar in 500.00 ml of water? fix this slide for ang Mol sugar = 75.5 g impure 85 g pure sug 100. g impure 1 mol sug 342.29 g sug = 0.18748 mol sug M sugar = mol pure sugar total volume solution =. 0.18748 0.50000 L = 0.375

What does a 3.5 M FeCl3 mean? 3.5 M FeCl3 implies: (1) a homogeneous solution of 3.5 moles of dry 100% pure FeCl3 dissolved in 1.00 Liter total solution volume. (3) Note the FeCl3 is not dissolved in 1.00 liter of water! (4) [Fe 3+ ] = 3.5M and [Cl - ] = 3 x 3.5 M (5) It can be used as a conversion factor

We can calculate concentrations of ions in solutions if the molarity is known. What is the molarity of each ion in 0.20M Al2(SO4) 3?

Preparing Solutions in the Laboratory A Weigh the solid needed. Transfer the solid to a volumetric flask that contains about half the final volume of solvent. B Dissolve the solid thoroughly by swirling. C Add solvent until the solution reaches its final volume.

Preparing Solutions in the Laboratory

How many grams of KI is required to make 500. ml of a 2.80 M KI solution? M KI M KI volume KI moles KI grams KI grams KI = 1 L 2.80 mol KI 166 g KI 500. ml x x x = 232 g KI 1000 ml 1 L soln 1 mol KI

Hydrochloric acid is sold commercially as a 12.0 M solution. How many moles of HCl are in 300.0 ml of 12.0 M solution? mol HCl = 1 L 12.0 mol HCl 300.0 ml = 3.60 mol HCl 1000 ml 1 L soln

Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M.

How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate? g of Na2HPO4 =? = 1.75 L soln X 0.460 moles Na2HPO4 1 L soln X 141.96 g Na 2 HPO 4 1 mol Na 2 HPO 4 = 114 g Na 2 HPO 4

How would you prepare 60.0 ml of 0.2 M HNO 3 from a stock solution of 4.00 M HNO 3? M i V i = M f V f M i = 4.00 M f = 0.200 V f = 0.06 L V i =? L V i = M f V f M i 0.200 M x 0.060 L = 4.00 M HNO3 = 0.003 L = 3.0 ml 3 ml of acid + 57 ml of water= 60.0 ml of solution

If 10.0 ml of 12.0 M HCl is added to enough water to give 100. ml of solution, what is the concentration of the solution?

We can apply the concepts of stoichiometry and molarity to chemical reactions that occur in solvents and solutions. Key is balanced equation! HCl(aq) + NaHCO3(aq) => NaCl(aq) + H2O(l) + CO2(g) Volume of HCl Molarity HCl Volume of NaHCO3 Molarity NaHCO3 Mole to Mole ratio

How many milliliters of 0.125 M solution are needed to neutralize 18.0 ml of 0.100 M HCl? HCl(aq) + NaHCO3(aq) => NaCl(aq) + H2O(l) + CO2(g)

Stoichiometry in Solution Specialized cells in the stomach release HCl to aid in digestion. If too much acid is released, the excess can cause heartburn and is sometimes neutralized with antacids. A common antacid is magnesium hydroxide, which reacts with the HCl to form water and magnesium chloride solution. As a government chemist testing commercial antacids, suppose you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of stomach acid react with a tablet containing 0.10g of magnesium hydroxide?

How many liters of stomach acid react with a tablet containing 0.10g of magnesium hydroxide? 1. Always translate nomenclature to chemical equation Mg(OH) 2 (s) + HCl(aq) MgCl 2 (aq) + H 2 O(l) unbalanced Mg(OH) 2 (s) + 2HCl(aq) MgCl 2 (aq) + 2H 2 O(l) balanced 2. Work in moles and use the stoichiometric factors. L HCl =? = 0.10 g Mg(OH) 2 1 mol Mg(OH) 2 58.33 g Mg(OH) 2 2 mol HCl 1 L HCl 1 mol Mg(OH) 2 0.1 mol HCl = = 3.4 x 10-2 L HCl

Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate, according to the equation below. How many milliliters of 0.125 M NaHCO3 solution are needed to neutralize 18.0 ml of 0.100 M HCl? HCl(aq) + NaHCO3(aq) => NaCl(aq) + H2O(l) + CO2(g) # mol HCl = MHCl x L = 0.100M HCl X 0.018L = 1.80 X 10-3 mol HCl # mol NaHCO3 = = 1.80 10 3 mol HCl 1 mol NaHCO 3 1 miol HCl 1 L solution 0.125 M NaHCO 3 =.0144 L solution

Limiting-Reactant Problems in Solution Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Suppose 0.050L of 0.010M mercury(ii) nitrate reacts with 0.020L of 0.10M sodium sulfide forming mercury(ii) sulfide. How many grams of mercury(ii) sulfide can form?

Suppose 0.050L of 0.010M mercury(ii) nitrate reacts with 0.020L of 0.10M sodium sulfide forming mercury(ii) sulfide. How many grams of mercury(ii) sulfide can form? Hg(NO 3 ) 2 (aq) + Na 2 S(aq) HgS(s) + 2NaNO 3 (aq) g HgS = 0.050 L Hg(NO 3 ) 2 0.010 mol Hg(NO 3 ) 2 1 L Hg(NO 3 ) 2 1 mol HgS 1 mol Hg(NO 3 ) 2 232.7 g HgS 1 mol HgS = 0.12 g HgS g HgS = 0.020 L Na 2 S 0.10 mol Na 2 S 1 L Na 2 S 1 mol HgS 1 mol Na 2 S 232.7 g HgS 1 mol HgS = 0.47 g HgS Hg(NO 3 ) 2 (aq) is the limiting reagent and 0.12 g HgS can be produced from this reaction.