Mass Relationships in Chemical Reactions. Chapter 3 Chang & Goldsby Modified by Dr. Juliet Hahn

Transcription

1 Mass Relationships in Chemical Reactions Chapter 3 Chang & Goldsby Modified by Dr. Juliet Hahn

2 Example 3.6 (3) We now write, 6.07 g CH 4 1 mol CH g CH 4 = mol CH 4 Thus, there is mole of CH 4 in 6.07 g of CH 4. end 9 am & 10 am class 8/25 F Check Should 6.07 g of CH 4 equal less than 1 mole of CH 4? What is the mass of 1 mole of CH 4?

3 Example 3.7 How many hydrogen atoms are present in 25.6 g of urea [(NH 2 ) 2 CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea is g. urea

4 Example 3.7 (1) Strategy We are asked to solve for atoms of hydrogen in 25.6 g of urea. We cannot convert directly from grams of urea to atoms of hydrogen. How should molar mass and Avogadro s number be used in this calculation? How many moles of H are in 1 mole of urea?

5 Example 3.7 (2) Solution To calculate the number of H atoms, we first must convert grams of urea to moles of urea using the molar mass of urea. This part is similar to Example 3.2. The molecular formula of urea shows there are four moles of H atoms in one mole of urea molecule, so the mole ratio is 4:1. Finally, knowing the number of moles of H atoms, we can calculate the number of H atoms using Avogadro s number. We need two conversion factors: molar mass and Avogadro s number.

6 Example 3.7 (3) We can combine these conversions grams of urea moles of urea moles of H atoms of H into one step: 25.6 gg NH 2 2 CO 1 mmmmmm NH 2 2CO gg NH 2 2 CO 4 mmmmmm HH 1 mmmmmm NH 2 2 CO HH aaaaaaaaaa 1 mmmmmm HH Check Does the answer look reasonable? = H atoms How many atoms of H would g of urea contain?

7 Formula Mass Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. NaCl 1Na 1Cl NaCl amu amu amu For any ionic compound formula mass amu = molar mass (grams) 1 formula unit NaCl = amu 1 mole NaCl = g NaCl 7

8 START go back after Exam I: Go back to Chapter 2 (the part about acids) before continuing with Chapter 3 (because too much memorization for time left before Exam I) 8

9 Come back later after Quiz I, cover chapter 3 and then come back to cover acid naming (because that is just memorization & not enough time before exam to memorize)

10 Acids An acid can be defined as a substance that yields hydrogen ions H + when dissolved in water. For example: HCl gas and HCl in water Pure substance, hydrogen chloride Dissolved in water (H 3 O + and Cl ), hydrochloric acid 10

11 Some Simple Acids Table 2.5 Some Simple Acids Acid Corresponding Anion HF (hydrofluoric acid) * F (fluoride) memorize the ones HCl (hydrochloric acid)* Cl (chloride) with the *, name HBr (hydrobromic acid)* Br (bromide) formula and name HI (hydroiodic acid)* I (iodide) HCN (hydrocyanic acid) CN (cyanide) H 2 S (hydrosulfuric acid) S 2 (sulfide) hydro ic acid all halogen group acids 11

12 Oxoacids An oxoacid is an acid that contains hydrogen, oxygen, and another element. (memorize the ones with the *) HNO 3 * H 2 CO 3 * nitric acid carbonic acid H 3 PO 4 * phosphoric acid H 2 SO 4 * sulfuric acid 12

13 Bases A base can be defined as a substance that yields hydroxide ions OH when dissolved in water. (named normally as polyatomic hydroxide ionic compound) NaOH KOH Ba OH 2 sodium hydroxide potassium hydroxide barium hydroxide 13

14 Hydrates (don t worry about these for now) Hydrates are compounds that have a specific number of water molecules attached to them. BaCl 2 2H 2 O LiCl H 2 O MgSO 4 7H 2 O Sr NO H 2 O barium chloride dihydrate lithium chloride monohydrate magnesium sulfate heptahydrate strontium nitrate tetrahydrate CuSO 4 5H 2 O CuSO 4 14

15 Common and Systematic Names of Compounds (not responsible for common names for now) Copyright McGraw-Hill Education. Permission required for reproduction or display. Table 2.7 Common and Systematic Names of Some Compounds Formula Common Name Systematic Name H 2 O Water Dihydrogen monoxide NH 3 Ammonia Trihydrogen nitride CO 2 Dry ice Solid carbon dioxide NaCl Table salt Sodium chloride N 2 O Laughing gas Dinitrogen monoxide CaCO 3 Marble, chalk, limestone Calcium carbonate CaO Quicklime Calcium oxide Ca(OH) 2 Slaked lime Calcium hydroxide NaHCO 3 Baking soda Sodium hydrogen carbonate Na 2 CO 3. 10H 2 O Washing soda Sodium carbonate decahydrate MgSO 4. 7H 2 O Epsom salt Magnesium sulfate heptahydrate Mg(OH) 2 Milk of magnesia Magnesium hydroxide CaSO 4. 2H 2 O Gypsum Calcium sulfate dehydrate 15

16 Organic Chemistry Organic chemistry is the branch of chemistry that deals with carbon compounds. (not responsible for now) H Functional Groups: H H C OH H C NH 2 H C C OH H O H methanol H methylamine H acetic acid 16

17 End go back after Exam I Go back to Chapter 2 (the part about acids) before continuing with Chapter 3 17

18 Percent Composition Percent composition of an element in a compound = n molar mass of element molar mass of compound 100% n is the number of moles of the element in 1 mole of the compound C 2 H 6 O 2 (12.01 g) %C = 100% = 52.14% g 6 (1.008 g) %H = 100% = 13.13% g 1 (16.00 g) %O = 100% = 34.73% 46.07g 52.14% % % = 100.0% 18

19 Example 3.8 Phosphoric acid (H 3 PO 4 ) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a tangy flavor. Calculate the percent composition by mass of H, P, and O in this compound. H 3 PO 4

20 Example 3.8 (1) Strategy Recall the procedure for calculating a percentage. Assume that we have 1 mole of H 3 PO 4. The percent by mass of each element (H, P, and O) is given by the combined molar mass of the atoms of the element in 1 mole of H 3 PO 4 divided by the molar mass of H 3 PO 4, then multiplied by 100 percent.

21 Example 3.8 (2) Solution The molar mass of H 3 PO 4 is g. The percent by mass of each of the elements in H 3 PO 4 is calculated as follows: %H = g H g H 3 PO 4 100% = %P = g P g H 3 PO 4 100% = 31.61% %O = g O g H 3 PO 4 100% = 65.31% Check Do the percentages add to 100 percent? The sum of the percentages is 3.086% % % = %. The small discrepancy from 100 percent is due to the way we rounded off.

22 Percent Composition and Empirical Formulas 22

23 Example 3.9 Ascorbic acid (vitamin C) cures scurvy. It is composed of percent carbon (C), 4.58 percent hydrogen (H), and percent oxygen (O) by mass. Determine its empirical formula. End 8/30 9 am class

24 Example 3.9 (1) Strategy In a chemical formula, the subscripts represent the ratio of the number of moles of each element that combine to form one mole of the compound. How can we convert from mass percent to moles? If we assume an exactly 100-g sample of the compound, do we know the mass of each element in the compound? How do we then convert from grams to moles?

25 Example 3.9 (2) Solution If we have 100 g of ascorbic acid, then each percentage can be converted directly to grams. In this sample, there will be g of C, 4.58 g of H, and g of O. Because the subscripts in the formula represent a mole ratio, we need to convert the grams of each element to moles. The conversion factor needed is the molar mass of each element. Let n represent the number of moles of each element so that n C = g C 1 mol C = mol C g C n H = g H 1 mol H = 4.54 mol H g H n O = g O 1 mol O = mol O g O

26 Example 3.9 (3) Thus, we arrive at the formula C H 4.54 O 3.406, which gives the identity and the mole ratios of atoms present. However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing all the subscripts by the smallest subscript (3.406): C: H: = 1.33 O: = 1 where the sign means approximately equal to. This gives CH 1.33 O as the formula for ascorbic acid. Next, we need to convert 1.33, the subscript for H, into an integer.

27 Example 3.9 (4) This can be done by a trial-and-error procedure: = = = 3.99 < 4 Because gives us an integer (4), we multiply all the subscripts by 3 and obtain C 3 H 4 O 3 as the empirical formula for ascorbic acid. End class 8/30/17W Check Are the subscripts in C 3 H 4 O 3 reduced to the smallest whole numbers?

28 Chemical Reactions and Equations A process in which one or more substance(s) is changed into one or more new substance(s) is a chemical reaction. A chemical equation uses chemical symbols to show what happens during a chemical reaction: reactants products 3 ways of representing the reaction of H 2 with O 2 to form H 2 O 28

29 How to Read Chemical Equations 2Mg + O 2 2MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg grams O 2 makes 80.6 g MgO NOT 2 grams Mg + 1 grams O 2 makes 2 g MgO 29

30 Balancing Chemical Equations 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 +O 2 CO 2 +H 2 O 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C 2 H 6 NOT C 4 H 12 30

31 Balancing Chemical Equations (1) 3. 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 End 9 am class 9/1 F 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3 C 2 H 6 + O 2 2CO 2 + 3H 2 O 31