Question 3.6: Calculating the Mass Percent of Each Element in a Compound f......
Calculating the Mass Percent of Each Element in a Compound from the Formula
Problem The effectiveness of fertilizers depends on their nitrogen content. Ammonium nitrate is a common fertilizer. What is the mass percent of each element in ammonium nitrate?
Plan We know the relative amounts (mol) of the elements from the formula, and we have to find the mass % of each element. We multiply the amount of each element by its molar mass to find its mass. Dividing each element’s mass by the mass of 1 mol of ammonium nitrate gives the mass fraction of that element, and multiplying the mass fraction by 100 gives the mass %. The calculation steps for any element (X) are shown in the road map (Fig 3.6).
Solution The formula is NH_4NO_3 (see Table 2.5). In 1 mol of NH_4NO_3, there are 2 mol of N, 4 mol of H, and 3 mol of O.
Converting amount (mol) of N to mass (g): We have 2 mol of N in 1 mol of NH_4NO_3, so
\text{Mass (g) of N} = 2 {\text{mol N}} \times \frac{14.01 g N}{1 \text{mol N}} = 28.02 g NCalculating the mass of 1 mol of NH_4NO_3:
ℳ = (2 × ℳ \text{of} N) + (4 × ℳ \text{of} H) + (3 × ℳ \text{of} O)
= (2 × 14.01 \text{g/mol N}) + (4 × 1.008 \text{g/mol H}) + (3 × 16.00 \text{g/mol O})
= 80.05 \text{g/mol} NH_4NO_3
Finding the mass fraction of N in NH_4NO_3:
\text{Mass fraction of N} = \frac{\text{total mass of N}}{\text{mass of 1 mol} NH_4NO_3} = \frac{28.02 g N}{80.05 g NH_4NO_3} = 0.3500Changing to mass %:
\text{Mass \% of N} = \text{mass fraction of} N × 100 = 0.3500 × 100
= 35.00 \text{mass \% N}
Combining the steps for each of the other elements in NH_4NO_3:
\text{Mass \% of H} = \frac{\text{mol H × }ℳ\text{ of H}}{\text{mass of 1 mol} NH_4NO_3} \times \frac{4 \cancel{\text{mol H}} \times \frac{1.008 g H}{1 \cancel{\text{mol H}}}}{80.05 g NH_4NO_3} \times 100
= 5.037 \text{mass \% H}
\text{Mass \% of O} = \frac{\text{mol O × }ℳ\text{ of O}}{\text{mass of 1 mol} NH_4NO_3} \times \frac{3 \cancel{\text{mol H}} \times \frac{16.00 g O}{1 \cancel{\text{mol O}}}}{80.05 g NH_4NO_3} \times 100
= 59.96 \text{mass \% O}
Check The answers make sense. The mass % of O is greater than that of N because there are more moles of O in the compound and the molar mass of O is greater. The mass % of H is small because its molar mass is small. The sum of the mass percents is 100.00%.
Comment From here on, you should be able to determine the molar mass of a compound, so that calculation will no longer be shown.
Table 2.5 Common Polyatomic Ions^*
| Formula | Name |
| Cations | |
| NH_4^+ | ammonium |
| H_3O^+ | hydronium |
| Anions | |
| CH_3COO^- (or C_2H_3O_2^-) | acetate |
| CN^- | cyanide |
| OH^- | hydroxide |
| ClO^- | hypochlorite |
| ClO_2^- | chlorite |
| ClO_3^- | chlorate |
| ClO_4^- | perchlorate |
| NO_2^- | nitrite |
| NO_3^- | nitrate |
| MnO_4^- | permanganate |
| CO_3^{2-} | carbonate |
| HCO_3^{-} | hydrogen carbonate (or bicarbonate) |
| CrO_4^{2-} | chromate |
| Cr_2O_7^{2-} | dichromate |
| O_2^{2-} | peroxide |
| PO_4^{3-} | phosphate |
| HPO_4^{2-} | hydrogen phosphate |
| H_2PO_4^{-} | dihydrogen phosphate |
| SO_3^{2-} | sulfite |
| SO_4^{2-} | sulfate |
| HSO_4^{-} | hydrogen sulfate (or bisulfate) |
^*Boldface ions are the most common.
