Chapter 3 Molecules, Compounds, and Chemical Equations

Chapter 3 Molecules, Compounds, and Chemical Equations

3.7 Formula Mass versus Molar mass Formula mass The average mass of a molecule or formula unit in amu also known as molecular mass or molecular weight (MW) whole = sum of the parts! Molar mass Total mass of a compound in gram per 1 mol of its molecules or formula unit

Molar Mass of Compounds the relative masses of molecules can be calculated from atomic masses Formula Mass = 1 molecule of H 2 O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu Why multiplying 2? since 1 mole of H 2 O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H 2 O = 2(1.01 g H) + 16.00 g O = 18.02 g so the Molar Mass of H 2 O is 18.02 g/mole 3

Molar Mass of Na 2 SO 4 Calculate the molar mass of Na 2 SO 4. Element Number of Moles Atomic Mass Total Mass in each element Na S O Total mass in 1 mol Na 2 SO 4 4

3.8 Mass Percent Composition Percentage of each element in a compound By mass Can be determined from 1. the formula of the compound 2. the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding Percentage part whole 100% 5

Example - Mass Percent as a Conversion Factor Calculate the mass percent of Na in NaCl Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?

Chemical Formulas and Elemental Composition chemical formulas have inherent in them relationships between numbers of atoms and molecules or moles of atoms and molecules these relationships can be used to convert between amounts of constituent elements and molecules like percent composition Vol A Grams A Moles A Moles B Grams B Grams A Moles A Moles B Grams B 7

Example Butane (C 4 H 10 ) is the liquid fuel in lighters. a. Determine the number of atoms ratio between carbon and 1 molecule of C 4 H 10 b. Determine the number moles ratio between C and 1 mol of C 4 H 10 c. How many grams of carbon are present within a lighter containing 7.5 ml of butane? The density of quid butane is 0.601 g/ml

Empirical Formula simplest, whole-number ratio of the atoms of elements in a compound can be determined from elemental analysis masses of elements formed when decompose or react compound combustion analysis percent composition 9

Steps in determine the Empirical formula Step 1: Obtain the mass of each element (in grams) E.G 100% = 100g therefore mass percent is the same numerical value in grams Step 2: Determine the numbers moles of each atom present Use molar mass of each element Step 3: Divide the smallest moles by numbers moles of each atom to obtain the closet integer as possible. if result is within 0.1 of whole number, round to whole number Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc then multiply with a factor to get the nearest integer as possible. E.g 1.5 x 2 = 3.0 atoms 1.33 x 3 = 3. 99 = 4 atoms Step 5: Write the result (number atoms) from step 4 as a subscript for the appropriate element.

Example Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70g/mol) and the rest fluorine (19.00 g/mol) An unknown sample gives the following mass percent: 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula?

Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound Multiple (n) = molecular mass empirical formula mass Molecular formula = empirical formula x n where n = 1, 2, 3, 4 12

Example Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula and molecular formula C = 60.00% H = 4.48% O = 35.53% 13

Determining Empirical Formulas: Elemental Analysis Combustion Analysis: A compound of unknown composition (containing a combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to produce the volatile combustion products CO 2 and H 2 O, which are separated and weighed by an automated instrument called a gas chromatograph. hydrocarbon + O 2 (g) xco 2 (g) + yh 2 O(g) carbon hydrogen

Combustion Analysis Unknown formula: CxHyOx (Oxygen can be replaced with other nonmetal) gco 2 moles CO 2 moles C gc gh 2 O moles H 2 O moles H gh g O = g sample (g H + g C) go moles O Follow steps in determine the empirical formula and molecular formula

Example Combustion of a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO 2 = 2.445 g H 2 O = 0.6003 g Determine the empirical formula of the compound 16

Example Upon combustion, a compound containing only carbon and hydrogen produced 1.60g CO 2 and 0.819g H 2 O. Find the empirical formula 17

Chemical Reactions Reactions involve chemical changes in matter resulting in new substances Reactions involve rearrangement and exchange of atoms to produce new molecules Elements are not transmuted during a reaction 18

Chemical Equations A chemical equation gives the formulas of the reactants on the left of the arrow. the formulas of the products on the right of the arrow. Reactants Product C(s) O 2 (g) CO 2 (g) 19

Symbols Used in Equations Symbols in chemical equations show TABLE the states of the reactants. the states of the products. the reaction conditions. 20

Chemical Equations are Balanced In a balanced chemical reaction no atoms are lost or gained. the number of reacting atoms is equal to the number of product atoms. 21

Balancing Chemical Equations A balanced chemical equation shows that the law of conservation of mass is adhered to. In a balanced chemical equation, the numbers and kinds of atoms on both sides of the reaction arrow are identical. 2Na(s) + Cl 2 (g) 2NaCl(s) left side: right side: 2 Na 2 Cl 2 Na 2 Cl

Balancing Chemical Equations 1. Write the unbalanced equation using the correct chemical formula for each reactant and product. H 2 (g) + O 2 (g) 2. Find suitable coefficients the numbers placed before formulas to indicate how many formula units of each substance are required to balance the equation. 2H 2 (g) + O 2 (g) H 2 O(l) 2H 2 O(l) 3. Reduce the coefficients to their smallest whole-number values, if necessary, by dividing them all by a common denominator. 2H 2 (g) + O 2 (g) 2H 2 O(l)

Balancing Chemical Equations 4. Check your answer by making sure that the numbers and kinds of atoms are the same on both sides of the equation. 2H 2 (g) + O 2 (g) left side: 4 H 2 O 2H 2 O(l) right side: 4 H 2 O

Balancing Chemical Equations Do not change subscripts when you balance a chemical equation. You are only allowed to change the coefficients. H 2 (g) + O 2 (g) H 2 O(l) unbalanced 2H 2 (g) + O 2 (g) 2H 2 O(l) H 2 (g) + O 2 (g) H 2 O 2 (l) Balanced properly Chemical equation changed!

Examples Balance the coefficients from reactants to products. A. N 2 (g) + H 2 (g) NH 3 (g) A. B. Co 2 O 3 (s) + C(s) Co(s) + CO 2 (g) Write a balanced equation for the reaction between a. carbon dioxide gas and aqueous potassium hydroxide to form potassium carbonate and water. b. The combustion of gaseous ethane (C 2 H 6 ) 26