• Essay Question 1 - Phosphoric acid (H₃PO₄) is used in detergents, fertilizers, toothpastes and flavoring in carbonated beverages. Calculate the percent composition of H₃PO₄. Ensure all work is shown.
• Essay Question 2 - Caffeine contains by mass 49.5% C, 5.15% H, 28.9% N and 16.5% O, and the molecular mass is 195g/mol. What are the empirical and molecular formulas of caffeine?  Ensure all work is shown.

 

Essay Question 1 - Phosphoric acid (H3PO4) is used in detergents, fertilizers, toothpastes and flavoring in carbonated beverages. Calculate the percent composition of H3PO4. Ensure all work is shown.

Percent composition is found by taking the ratio between the mass of an element in a compound to the mass of the entire compound. We can calculate the percent composition of each element by first calculating the molar mass of H₃PO₄ and then seeing what percent of this total each element makes up.

H₃PO₄: 3(1.01) + 30.97 + 4(16.0) = 98.0 g/mol

%H: (3.03 g/98.0 g) x 100 = 3.09% H

%P: (30.97 g/98.0 g) x 100 = 31.60% P

%O: (64.0 g/98.0 g) x 100 = 65.31% O

Essay Question 2 - Caffeine contains by mass 49.5% C, 5.15% H, 28.9% N and 16.5% O, and the molecular mass is 195g/mol. What are the empirical and molecular formulas of caffeine?  Ensure all work is shown.

We can determine the empirical formula by treating the given percentages as masses. We then need to convert the mass of each element to moles by using their molar mass as a conversion factor. If we then take the mole ratio between all of the atoms in the compound, we will obtain the subscripts in the empirical formula.

49.5 g C x (1 mol C/12.01 g C) = 4.122... mol C

5.15 g H x (1 mol H/1.01 g H) = 5.099... mol H

28.9 g N x (1 mol N/14.01 g N) = 2.063... mol N

16.5 g O x (1 mol O/16.0 g O) = 1.03125 mol O

Ratio of Moles C : H : N : O = 4 : 5 : 2 : 1

The empirical formula is therefore C₄H₅N₂O

The molecular formula is found by taking the ratio of the molecular mass to the empirical formula mass. This ratio will be a whole number that we then multiply each of the subscripts in the empirical formula by.

Empirical Formula Mass: 4(12.01) + 5(1.01) + 2(14.01) + 16 = 97.11 g/mol

Molecular Mass to Empirical Formula Mass = 195/97.11 = 2

This indicates the subscripts of the empirical formula must be doubled.

The molecular formula is C₈H₁₀N₄O₂