Question

what number of molecules (or formula units) are present in 6.65 g of each of the following compounds?
a.
1.334x10^23 molecules
b.
1.334x10^23 molecules
c. $(nh_4)_2cr_2o_7$
1.589x10^22 formula units
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Explanation:

Step1: Find molar mass of NH₃ (a)

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

Step2: Calculate moles of NH₃

$n(\text{NH}_3) = \frac{m}{M} = \frac{6.65}{17.034} \approx 0.3904\ \text{mol}$

Step3: Find number of NH₃ molecules

Use $N = n\times N_A$, $N_A=6.022\times10^{23}\ \text{mol}^{-1}$
$N(\text{NH}_3) = 0.3904\times6.022\times10^{23} \approx 2.35\times10^{23}\ \text{molecules}$
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Step1: Find molar mass of N₂H₄ (b)

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

Step2: Calculate moles of N₂H₄

$n(\text{N}_2\text{H}_4) = \frac{6.65}{32.052} \approx 0.2075\ \text{mol}$

Step3: Find number of N₂H₄ molecules

$N(\text{N}_2\text{H}_4) = 0.2075\times6.022\times10^{23} \approx 1.25\times10^{23}\ \text{molecules}$
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Step1: Find molar mass of (NH₄)₂Cr₂O₇ (c)

Molar mass: $M((\text{NH}_4)_2\text{Cr}_2\text{O}_7) = 2\times(14.01+4\times1.008) + 2\times52.00 + 7\times16.00 = 252.076\ \text{g/mol}$

Step2: Calculate moles of (NH₄)₂Cr₂O₇

$n((\text{NH}_4)_2\text{Cr}_2\text{O}_7) = \frac{6.65}{252.076} \approx 0.02638\ \text{mol}$

Step3: Find number of formula units

$N((\text{NH}_4)_2\text{Cr}_2\text{O}_7) = 0.02638\times6.022\times10^{23} \approx 1.59\times10^{22}\ \text{formula units}$

Answer:

a. $\boldsymbol{2.35\times10^{23}}$ molecules
b. $\boldsymbol{1.25\times10^{23}}$ molecules
c. $\boldsymbol{1.59\times10^{22}}$ formula units