Calculating Empirical Formula from % Composition

• Empirical formulas contain the information needed to calculate the percent composition of compounds since they preserve the ration of atoms in a compound. When the mole ratio is conserved, so is the mass ratio. 
• Scientists use this to determine the empirical formula of a compound by experimentally determining the percent composition. 
• Suppose a chemist is given 100.0 g of an unknown substance. After analysis, he concludes that it contains 75.00 g (75%) C and 25.00 g (25%) O. We are able to figure out the empirical formula from this information.
• 75.00 g C x 1 mol C/12.01 g C = 6.245 mol C
• 25.00 g H x 1 mol H/1.01 g H = 24.80 mol H
• Next to get a 1: something ratio, we divide both numbers by the smallest.
• 6.245/6.245 = 1
• 24.80/6.245 = 3.97 (this gets rounded to 4)
• with a ratio of 1:4, we can write our formula as C₁H₄ or CH₄
• A laboratory analysis of an unknown gas has determined that the bass is 72.55% oxygen and 27.45% carbon by mass. What is the empirical formula of the compound?
• 73.55 g O x 1 mol O/16.00 g O = 4.534 mol O
• 27.45 g C x 1 mol g C/12.01 g C = 2.286 mol C
• 4.543/2.286 : 2.286/2.286 = 1.983:1.000
• Giving a ratio of 2 O:1 C
• CO₂
• A 5.000 g sample of an unknown compound contains 1.844 g of N and 3.156 g of O. Find the empirical formula.
• 1.844 g N x 1 mol N/14.01 g N = 0.1316 mol N
• 3.156 g O x 1 mol O/16.00 g O = 0.1973 mol O
• 0.1316/0.1316 : 0.1973/0.1316 = 1:1.499
• Because we do not have half numbers in empirical formulas, we must eliminate the fraction. Making this 2:3
• N₂O₃
• Caffeine, in coffee and some carbonated beverages, is 5.170% H, 16.49% O, 28.86% N, and 49.48% C by mass. What is the empirical formula of caffeine? (C₄H₅N₂O)