Answer
a - Mass percentage of C = 42.9 %
Mass percentage of O = 57.1 %
b- Mass percentage of C = 60.0 %
Mass percentage of O = 40 %
c- Mass percentage of Na = 19.2 %
Mass percentage of H= 1.67 %
Mass percentage of P= 25.8 %
Mass percentage of O = 53.3 %
d- Mass percentage of Co= 32.2 %
Mass percentage of N= 15.3 %
Mass percentage of O = 52.5 %
Work Step by Step
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 - To find the mass percentage of C and O in CO, first find the Molar mass of CO.
Molar mass of CO = 1 x 12.0 + 1 x 16.0
Molar mass of CO = 28 g/mol ( to three significant figures)
Based on the chemical formula CO, in 1 mole of CO , there is 1 mole of C, 1 mole of O.
Mass percentage of C = $\frac{mass_(1 C mole)}{Molar mass_(CO)}$$\times$ 100%
Mass percentage of C = $\frac{12.0}{28.0}$$\times$ 100%
Mass percentage of C = 42.9 %
Mass percentage of O = $\frac{mass_(1 O mole)}{Molar mass_(CO)}$$\times$ 100%
Mass percentage of O = $\frac{16.0}{28.0}$$\times$ 100%
Mass percentage of O = 57.1 %
b - $CO_{2}$. Molar mass of $CO_{2}$.
Molar mass of $CO_{2}$ = 2 x 12.0 + 1 x 16.0
Molar mass of $CO_{2}$ = 40.0 g/mol ( to three significant figures)
Based on the chemical formula $CO_{2}$, in 1 mole of $CO_{2}$, there re 2 mole of C, 1 mole of O.
Mass percentage of C = $\frac{mass_(2 C mole)}{Molar mass_(CO_{2})}$$\times$ 100%
Mass percentage of C = $\frac{2x12.0}{40.0}$$\times$ 100%
Mass percentage of C = 60.0 %
Mass percentage of O = $\frac{mass_(1 O mole)}{Molar mass_(CO_{2})}$$\times$ 100%
Mass percentage of O = $\frac{16.0}{40.0}$$\times$ 100%
Mass percentage of O = 40 %
c - $NaH_{2}PO_{4}$
Molar mass of $NaH_{2}PO_{4}$ = 1 x 23.0 + 2 x 1.00 +1 x 31.0 4x 16.0
Molar mass of $NaH_{2}PO_{4}$ = 120 g/mol ( to three significant figures)
Based on the chemical formula$NaH_{2}PO_{4}$, in 1 mole of $NaH_{2}PO_{4}$, there are 1 mole Na, 2 mole H, 1 mole of P and 4 mole O.
Mass percentage of Na = $\frac{mass_(1 Na mole)}{Molar mass_(NaH_{2}PO_{4})}$$\times$ 100%
Mass percentage of Na = $\frac{2 X 23.0}{120}$$\times$ 100%
Mass percentage of Na = 19.2 %
Mass percentage of H = $\frac{mass_(2 H mole)}{Molar mass_(NaH_{2}PO_{4})}$$\times$ 100%
Mass percentage of H = $\frac{2 X 1.00}{120}$$\times$ 100%
Mass percentage of H= 1.67 %
Mass percentage of P = $\frac{mass_(1 mole mole)}{Molar mass_(NaH_{2}PO_{4})}$$\times$ 100%
Mass percentage of P= $\frac{31.0}{120}$$\times$ 100%
Mass percentage of P= 25.8 %
Mass percentage of O = $\frac{mass_(4 O mole)}{Molar mass_(NaH_{2}PO_{4})}$$\times$ 100%
Mass percentage of O = $\frac{4X 16.0}{120.0}$$\times$ 100%
Mass percentage of O = 53.3 %
d - $Co(NO_{3})_2$
Molar mass of $Co(NO_{3})_2$ = 1 x 59.0 +2 x 14.0+ 6 x 16.0
Molar mass of $Co(NO_{3})_2$ = 183 g/mol ( to three significant figures)
Based on the chemical formula in 1 mole of $Co(NO_{3})_2$, there are 1 mole Co, 2 mole N and 6 mole O.
Mass percentage of Co= $\frac{mass_(1 Co mole)}{Molar mass_(Co(NO_{3})_2)}$$\times$ 100%
Mass percentage of Co= $\frac{59.0}{183}$$\times$ 100%
Mass percentage of Co= 32.2 %
Mass percentage of N= $\frac{mass_(2 N mole)}{Molar mass_(NaH_{2}PO_{4})}$$\times$ 100%
Mass percentage of N= $\frac{2X14.0}{183}$$\times$ 100%
Mass percentage of N= 15.3 %
Mass percentage of O = $\frac{mass_(6 O mole)}{Molar mass_(NaH_{2}PO_{4})}$$\times$ 100%
Mass percentage of O = $\frac{6X 16.0}{183}$$\times$ 100%
Mass percentage of O = 52.5 %
