QUESTION IMAGE
Question
find the percent compositions of all of the elements in the following compounds:
- cubr₂
cu: _______
br: _______
- naoh
na: _______
o: _______
h: _______
- (nh₄)₂s
n: _______
h: _______
s: _______
- n₂s₃
n: _______
s: _______
Problem 1: $\boldsymbol{CuBr_2}$
Step1: Calculate molar mass of $CuBr_2$
Molar mass of $Cu$ = 63.55 g/mol, $Br$ = 79.90 g/mol.
Molar mass = $63.55 + 2 \times 79.90 = 63.55 + 159.8 = 223.35$ g/mol.
Step2: Percent composition of $Cu$
Percent $Cu$ = $\frac{63.55}{223.35} \times 100 \approx 28.45\%$
Step3: Percent composition of $Br$
Percent $Br$ = $\frac{159.8}{223.35} \times 100 \approx 71.55\%$ (or $2 \times \frac{79.90}{223.35} \times 100$)
Problem 2: $\boldsymbol{NaOH}$
Step1: Calculate molar mass of $NaOH$
Molar mass of $Na$ = 22.99 g/mol, $O$ = 16.00 g/mol, $H$ = 1.008 g/mol.
Molar mass = $22.99 + 16.00 + 1.008 = 39.998 \approx 40.00$ g/mol.
Step2: Percent composition of $Na$
Percent $Na$ = $\frac{22.99}{40.00} \times 100 \approx 57.48\%$
Step3: Percent composition of $O$
Percent $O$ = $\frac{16.00}{40.00} \times 100 = 40.00\%$
Step4: Percent composition of $H$
Percent $H$ = $\frac{1.008}{40.00} \times 100 \approx 2.52\%$
Problem 3: $\boldsymbol{(NH_4)_2S}$
Step1: Calculate molar mass of $(NH_4)_2S$
Molar mass of $N$ = 14.01 g/mol, $H$ = 1.008 g/mol, $S$ = 32.07 g/mol.
Molar mass = $2 \times (14.01 + 4 \times 1.008) + 32.07 = 2 \times (14.01 + 4.032) + 32.07 = 2 \times 18.042 + 32.07 = 36.084 + 32.07 = 68.154$ g/mol.
Step2: Percent composition of $N$
Total $N$ mass = $2 \times 14.01 = 28.02$ g/mol.
Percent $N$ = $\frac{28.02}{68.154} \times 100 \approx 41.11\%$
Step3: Percent composition of $H$
Total $H$ mass = $8 \times 1.008 = 8.064$ g/mol.
Percent $H$ = $\frac{8.064}{68.154} \times 100 \approx 11.83\%$
Step4: Percent composition of $S$
Percent $S$ = $\frac{32.07}{68.154} \times 100 \approx 47.06\%$
Problem 4: $\boldsymbol{N_2S_2}$
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s:
- $Cu$: $\approx 28.45\%$, $Br$: $\approx 71.55\%$
- $Na$: $\approx 57.48\%$, $O$: $40.00\%$, $H$: $\approx 2.52\%$
- $N$: $\approx 41.11\%$, $H$: $\approx 11.83\%$, $S$: $\approx 47.06\%$
- $N$: $\approx 30.40\%$, $S$: $\approx 69.60\%$