Chapter 3 - Calculations with Chemical Formulas and Equations - Questions and Problems - Page 120: 3.58

a - Mass percentage of N = 30.4 % Mass percentage of O = 69.6 % b - Mass percentage of H = 6.7 % Mass percentage of N = 93.3 % c - Mass percentage of K = 28.2% Mass percentage of Cl = 25.6% Mass percentage of O = 46.2% d - Mass percentage of Mg= 16.4 % Mass percentage of N= 18.9 % Mass percentage of O = 64.7 %

Strategy: To calculate the mass percentage of elements in the compound : 1- calculate the Molar mass of the compound 2- Calculate the mass of each element to the Molar mass and express this in percentage. a - $NO_{2}$. Molar mass of $NO_{2}$. Molar mass of $NO_{2}$ = 1 x 14.0 + 2 x 16.0 Molar mass of $NO_{2}$ = 46.0 g/mol ( to three significant figures) Based on the chemical formula $NO_{2}$, in 1 mole of $NO_{2}$, there are 1 mole of N, 2 mole of O. Mass percentage of N = $\frac{mass_(1 N mole)}{Molar mass_(NO_{2})}$$\times$ 100% Mass percentage of N = $\frac{1\times14.0}{46.0}$$\times$ 100% Mass percentage of N = 30.4 % Mass percentage of O = $\frac{mass_(2 O mole)}{Molar mass_(NO_{2})}$$\times$ 100% Mass percentage of O = $\frac{2 \times16.0}{46.0}$$\times$ 100% Mass percentage of O = 69.6 % b - $H_{2}N_{2}$ Molar mass $H_{2}N_{2}$ = 2 x 1.00 + 2 x 14.00 Molar mass $H_{2}N_{2}$ =30.0 g/ mol Mass percentage of H = $\frac{mass_(2 mole H)}{Molar mass_(H_{2}N_{2})}$$\times$ 100% Mass percentage of H = $\frac{2\times1.00}{30}$$\times$ 100% Mass percentage of H = 6.7 % Mass percentage of N = $\frac{mass_(2 mole N)}{Molar mass_(H_{2}N_{2})}$$\times$ 100% Mass percentage of N = $\frac{2\times14.0}{30}$$\times$ 100% Mass percentage of N = 93.3 % c - $KClO_{4}$ Molar mass $KClO_{4}$ = 1 x 39.0 + 1 x 35.5 + 4 x 16.0 Molar mass $KClO_{4}$ =138.5 g/mol Mass percentage of K =$\frac{mass_(1 mole K)}{Molar mass_(KClO_{4})}$$\times$ 100% Mass percentage of K = $\frac{1\times39.0}{138.5}$$\times$ 100% Mass percentage of K = 28.2% Mass percentage of Cl =$\frac{mass_(1 mole Cl)}{Molar mass_(KClO_{4})}$$\times$ 100% Mass percentage of Cl = $\frac{1\times35.5}{138.5}$$\times$ 100% Mass percentage of Cl = 25.6% Mass percentage of O =$\frac{mass_(4 mole O)}{Molar mass_(KClO_{4})}$$\times$ 100% Mass percentage of O = $\frac{4\times16.0}{138.5}$$\times$ 100% Mass percentage of O = 46.2% d - $Mg(NO_{3})_2$ Molar mass of $Mg(NO_{3})_2$ = 1 x 24.3 +2 x 14.0+ 6 x 16.0 Molar mass of $Mg(NO_{3})_2$ = 148.3 g/mol ( to three significant figures) Based on the chemical formula in 1 mole of $Mg(NO_{3})_2$, there are 1 mole Mg, 2 mole N and 6 mole O. Mass percentage of Mg= $\frac{mass_(1 Mg mole)}{Molar mass_(Mg(NO_{3})_2)}$$\times$ 100% Mass percentage of Mg= $\frac{24.3}{148.3}$$\times$ 100% Mass percentage of Mg= 16.4 % Mass percentage of N= $\frac{mass_(2 N mole)}{Molar mass_(Mg(NO_{3})_2)}$$\times$ 100% Mass percentage of N= $\frac{2X14.0}{148.3}$$\times$ 100% Mass percentage of N= 18.9 % Mass percentage of O = $\frac{mass_(6 mole O)}{Molar mass_(Mg(NO_{3})_2)}$$\times$ 100% Mass percentage of O = $\frac{6X 16.0}{148.3}$$\times$ 100% Mass percentage of O = 64.7 %