Question

what number of atoms of nitrogen are present in 6.65 g of each of the following compounds?
a. $\boldsymbol{square}$ atoms
b. $\boldsymbol{square}$ atoms
c. $ce{(nh4)2cr2o7}$ $\boldsymbol{square}$ atoms

Explanation:

Step1: Identify compound (a) formula

The molecule is $\text{NH}_3$ (1 N, 3 H atoms).

Step2: Calculate molar mass of (a)

Molar mass $M_{\text{NH}_3} = 14.01 + 3\times1.008 = 17.034\ \text{g/mol}$

Step3: Find moles of (a)

$\text{Moles} = \frac{\text{Mass}}{M} = \frac{6.65}{17.034} \approx 0.3904\ \text{mol}$

Step4: Calculate N atoms for (a)

1 mole $\text{NH}_3$ has 1 mole N atoms.
Number of N atoms $= 0.3904 \times 6.022\times10^{23} \approx 2.35\times10^{23}$
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Step1: Identify compound (b) formula

The molecule is $\text{N}_2\text{H}_4$ (2 N, 4 H atoms).

Step2: Calculate molar mass of (b)

Molar mass $M_{\text{N}_2\text{H}_4} = 2\times14.01 + 4\times1.008 = 32.052\ \text{g/mol}$

Step3: Find moles of (b)

$\text{Moles} = \frac{6.65}{32.052} \approx 0.2075\ \text{mol}$

Step4: Calculate N atoms for (b)

1 mole $\text{N}_2\text{H}_4$ has 2 moles N atoms.
Number of N atoms $= 0.2075 \times 2 \times 6.022\times10^{23} \approx 2.49\times10^{23}$
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Step1: Calculate molar mass of (c)

Molar mass $M_{(\text{NH}_4)_2\text{Cr}_2\text{O}_7} = 2\times(14.01+4\times1.008) + 2\times52.00 + 7\times16.00$
$= 2\times18.042 + 104.00 + 112.00 = 36.084 + 216.00 = 252.084\ \text{g/mol}$

Step2: Find moles of (c)

$\text{Moles} = \frac{6.65}{252.084} \approx 0.02638\ \text{mol}$

Step3: Calculate N atoms for (c)

1 mole $(\text{NH}_4)_2\text{Cr}_2\text{O}_7$ has 2 moles N atoms.
Number of N atoms $= 0.02638 \times 2 \times 6.022\times10^{23} \approx 3.18\times10^{22}$

Answer:

a. $\boldsymbol{2.35\times10^{23}}$ atoms
b. $\boldsymbol{2.49\times10^{23}}$ atoms
c. $\boldsymbol{3.18\times10^{22}}$ atoms