“Percent Composition and Chemical Formulas” Chemistry 10.3 “Percent Composition and Chemical Formulas”

I. Percent Composition A. Def – the percent by mass of each element in a compound. 1. Must = 100% 2. Formula… Grams of element X Grams of compound % mass of element X = X 100%

3. Ex: An 8. 20 g piece of magnesium combines with 5 3. Ex: An 8.20 g piece of magnesium combines with 5.40 g of oxygen to form a compound. What is the % composition of this compound? -mass of compound = % Mg = 8.20 g 13.60 g = 60.3% 13.60 g X 100%

% O = 5.40 g 13.60 g = 39.7% *Does this make sense? 39.7% + 60.3% = 100% X 100% Yes

4. Ex: 9. 03 g of Mg combine completely with 3 4. Ex: 9.03 g of Mg combine completely with 3.48 g of N to form a compound. What is the % composition of this compound? 5. 29.0 g Ag completely react with 4.30 g of S to form a compound. What is the % composition of this compound? 72.2 % = Mg 27.8 % = N 87.1 % = Ag 12.9 % = S

B. Percent Composition from the Chemical Formula 1. Formula %Mass = grams of element in 1 mol of compound molar mass of compound 2. Must = 100% 3. Add molar mass + # of moles 4. Ex: Calculate the % composition of propane (C3H8) X 100%

molar mass = % C = 36.0 44.0 % H = 8.0 44.1 g /mol X 100% % C = 81.8%

5. Calculate the % composition of the following compounds: a. C2H6 = b. NaHSO4 = c. NH4Cl = d. CO(NH2)2 = %C = 79.9 %H = 20.1 %Na = 19.1 %H = .8 %S = 26.7 %O = 53.3 %N = 26.2 %H = 7.6 %Cl = 66.3 %C = 20.0 %O = 26.6 %N = 46.6 %H = 6.7

II. Empirical Formalas A. Def – gives the lowest whole # ratio of atoms in a compound. 1. can be used for atoms or moles 2. Ex: HO  H2O2 (hydrogen peroxide) 3. Empirical formula can be the same as molecular formula. -Ex: CO2 (1:2 ratio) 4. What are the empirical formulas for the following? a. C6H12O6 b. C4H8 c. N4O10 CH2O CH2 N2O5

5. Ethyne (C2H2) also called acetylene, is a gas used in welder’s torches. Styrene (C8H8) is used in making polystyrene. -What is similar about these 2 compounds? both have the same empirical formula (CH)

B. Calculating Empirical Formulas 1. Ex: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? a. Step 1 – determine the # of moles (assume %’s are in grams) b. Step 2 – divide by smallest mole # c. Step 3 – Check for decimals (2.5, 3.75, 1.25) 2. Solve (step 1) 25.9 g N x 1 mol N 74.1 g O x 1 mol O = 1.85 mol N 14.0 g N = 4.63 mol O 16.0 g O

(Step 2) 1.85 mol N 1.85 4.63 mol O = 1 mol N = 2.5 mol O

(Step 3) *What can we multiply by to get a whole # ratio between N and O? = 2 1 mol N x 2 = N2 2.5 mol O x 2 = O5 Empirical Formula = N2O5

Check answer by finding % composition for N and O. Calculate the empirical formulas for the following… 1. 94.1% O, 5.9% H 2. 79.8% C, 20.2% H 3. 67.6% Hg, 10.8% S, 21.6 O 4. 27.59% C, 1.15% H, 16.09% N, 55.17% O OH CH3 HgSO4 C2HNO3

III. Calculating Molecular Formulas (includes ionic compounds) A. Determine empirical formula B. Determine molar mass of compound C. Divide… Molar Mass Emp. Formula Mass D. Take each element in the empirical formula times “step C” E. Write out new formula

Examples: 1. Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH4N.