Question

percent composition worksheet ii
find the percent compositions of all of the elements in the following compounds:

1. cubr₂

cu: 28%
br: 72%

1. naoh

na:
o:
h:

1. (nh₄)₂s

n:
h:
s:

1. n₂s₂

n:
s:

Explanation:

Step1: Recall Percent Composition Formula

The formula for percent composition of an element in a compound is $\text{Percent Composition} = \frac{\text{Mass of Element in Compound}}{\text{Molar Mass of Compound}} \times 100\%$.

Step2: Solve for $\ce{NaOH}$

Step 2.1: Calculate Molar Mass of $\ce{NaOH}$

Molar mass of $\ce{Na}$ is $22.99\ \text{g/mol}$, $\ce{O}$ is $16.00\ \text{g/mol}$, and $\ce{H}$ is $1.008\ \text{g/mol}$.
Molar mass of $\ce{NaOH} = 22.99 + 16.00 + 1.008 = 39.998\ \text{g/mol} \approx 40.00\ \text{g/mol}$.

Step 2.2: Percent Composition of $\ce{Na}$

Mass of $\ce{Na}$ in $\ce{NaOH}$ is $22.99\ \text{g/mol}$.
$\% \ce{Na} = \frac{22.99}{40.00} \times 100\% \approx 57.48\%$.

Step 2.3: Percent Composition of $\ce{O}$

Mass of $\ce{O}$ in $\ce{NaOH}$ is $16.00\ \text{g/mol}$.
$\% \ce{O} = \frac{16.00}{40.00} \times 100\% = 40.00\%$.

Step 2.4: Percent Composition of $\ce{H}$

Mass of $\ce{H}$ in $\ce{NaOH}$ is $1.008\ \text{g/mol}$.
$\% \ce{H} = \frac{1.008}{40.00} \times 100\% \approx 2.52\%$.

Step3: Solve for $\ce{(NH4)2S}$

Step 3.1: Calculate Molar Mass of $\ce{(NH4)2S}$

Molar mass of $\ce{N}$ is $14.01\ \text{g/mol}$, $\ce{H}$ is $1.008\ \text{g/mol}$, $\ce{S}$ is $32.07\ \text{g/mol}$.
There are $2\ \ce{N}$ atoms, $8\ \ce{H}$ atoms, and $1\ \ce{S}$ atom.
Molar mass of $\ce{(NH4)2S} = (2 \times 14.01) + (8 \times 1.008) + 32.07 = 28.02 + 8.064 + 32.07 = 68.154\ \text{g/mol} \approx 68.15\ \text{g/mol}$.

Step 3.2: Percent Composition of $\ce{N}$

Mass of $\ce{N}$ in $\ce{(NH4)2S}$ is $2 \times 14.01 = 28.02\ \text{g/mol}$.
$\% \ce{N} = \frac{28.02}{68.15} \times 100\% \approx 41.12\%$.

Step 3.3: Percent Composition of $\ce{H}$

Mass of $\ce{H}$ in $\ce{(NH4)2S}$ is $8 \times 1.008 = 8.064\ \text{g/mol}$.
$\% \ce{H} = \frac{8.064}{68.15} \times 100\% \approx 11.83\%$.

Step 3.4: Percent Composition of $\ce{S}$

Mass of $\ce{S}$ in $\ce{(NH4)2S}$ is $32.07\ \text{g/mol}$.
$\% \ce{S} = \frac{32.07}{68.15} \times 100\% \approx 47.06\%$.

Step4: Solve for $\ce{N2S2}$

Step 4.1: Calculate Molar Mass of $\ce{N2S2}$

Molar mass of $\ce{N}$ is $14.01\ \text{g/mol}$, $\ce{S}$ is $32.07\ \text{g/mol}$.
There are $2\ \ce{N}$ atoms and $2\ \ce{S}$ atoms.
Molar mass of $\ce{N2S2} = (2 \times 14.01) + (2 \times 32.07) = 28.02 + 64.14 = 92.16\ \text{g/mol}$.

Step 4.2: Percent Composition of $\ce{N}$

Mass of $\ce{N}$ in $\ce{N2S2}$ is $28.02\ \text{g/mol}$.
$\% \ce{N} = \frac{28.02}{92.16} \times 100\% \approx 30.40\%$.

Step 4.3: Percent Composition of $\ce{S}$

Mass of $\ce{S}$ in $\ce{N2S2}$ is $64.14\ \text{g/mol}$.
$\% \ce{S} = \frac{64.14}{92.16} \times 100\% \approx 69.60\%$.

Answer:

2) $\ce{NaOH}$

$\ce{Na}$: $\approx 57.48\%$
$\ce{O}$: $40.00\%$
$\ce{H}$: $\approx 2.52\%$

3) $\ce{(NH4)2S}$

$\ce{N}$: $\approx 41.12\%$
$\ce{H}$: $\approx 11.83\%$
$\ce{S}$: $\approx 47.06\%$

4) $\ce{N2S2}$

$\ce{N}$: $\approx 30.40\%$
$\ce{S}$: $\approx 69.60\%$