mole review worksheet\n1. how many moles are in 3.4 x 10^25 molecules of so₃?\n2. what is the molar mass of…

1. How many moles are in $3.4\times 10^{25}$ molecules of $SO_3$?

Explanation:

Step1: Recall Avogadro's number

Use the formula $n=\frac{N}{N_A}$, where $n$ is the number of moles, $N$ is the number of molecules and $N_A = 6.022\times 10^{23}\text{ molecules/mol}$.

Step2: Substitute values

$n=\frac{3.4\times 10^{25}\text{ molecules}}{6.022\times 10^{23}\text{ molecules/mol}}\approx 56.5\text{ mol}$

Answer:

$56.5\text{ mol}$

2. What is the molar mass of $C_3H_8$?

Explanation:

Step1: Find molar - mass of each element

The molar mass of $C$ is $12.01\text{ g/mol}$ and of $H$ is $1.01\text{ g/mol}$.

Step2: Calculate molar mass of $C_3H_8$

$M=(3\times12.01\text{ g/mol})+(8\times1.01\text{ g/mol}) = 36.03\text{ g/mol}+8.08\text{ g/mol}=44.11\text{ g/mol}$

Answer:

$44.11\text{ g/mol}$

3. What is the molar mass of $CaCl_2$?

Explanation:

Step1: Find molar - mass of each element

The molar mass of $Ca$ is $40.08\text{ g/mol}$ and of $Cl$ is $35.45\text{ g/mol}$.

Step2: Calculate molar mass of $CaCl_2$

$M = 40.08\text{ g/mol}+2\times35.45\text{ g/mol}=40.08\text{ g/mol}+70.9\text{ g/mol}=110.98\text{ g/mol}$

Answer:

$110.98\text{ g/mol}$

4. How many grams are in 12.50 moles of $K_2SO_4$?

Explanation:

Step1: Find molar mass of $K_2SO_4$

Molar mass of $K$ is $39.10\text{ g/mol}$, $S$ is $32.07\text{ g/mol}$ and $O$ is $16.00\text{ g/mol}$. $M=(2\times39.10\text{ g/mol})+32.07\text{ g/mol}+(4\times16.00\text{ g/mol})=78.2\text{ g/mol}+32.07\text{ g/mol}+64.0\text{ g/mol}=174.27\text{ g/mol}$

Step2: Use the formula $m = n\times M$

$m=12.50\text{ mol}\times174.27\text{ g/mol}=2178.375\text{ g}$

Answer:

$2178.375\text{ g}$

5. How many molecules are in 4.50 moles of $H_2O$?

Explanation:

Step1: Recall Avogadro's number

Use the formula $N=n\times N_A$, where $n = 4.50\text{ mol}$ and $N_A=6.022\times 10^{23}\text{ molecules/mol}$

Step2: Substitute values

$N = 4.50\text{ mol}\times6.022\times 10^{23}\text{ molecules/mol}=2.71\times 10^{24}\text{ molecules}$

Answer:

$2.71\times 10^{24}\text{ molecules}$

6. How many moles are in 11.94 g of $NaCl$?

Explanation:

Step1: Find molar mass of $NaCl$

Molar mass of $Na$ is $22.99\text{ g/mol}$ and of $Cl$ is $35.45\text{ g/mol}$. $M = 22.99\text{ g/mol}+35.45\text{ g/mol}=58.44\text{ g/mol}$

Step2: Use the formula $n=\frac{m}{M}$

$n=\frac{11.94\text{ g}}{58.44\text{ g/mol}}\approx0.204\text{ mol}$

Answer:

$0.204\text{ mol}$

7. Label the following as empirical or molecular

a. $C_2N_2H_8$: Molecular (can be simplified to $CH_4N$) b. $C_3H_8O$: Molecular (simplest ratio is the same as the given formula) c. $Be_2(Cr_2O_7)_2$: Molecular (simplest ratio is not given in this form) d. $Sb_4S_6$: Molecular (can be simplified to $Sb_2S_3$)

Answer:

a. Molecular b. Molecular c. Molecular d. Molecular

8. What is the percent composition of the compounds below

a. $Al(NO_3)_3$

Explanation:

Step1: Find molar mass of $Al(NO_3)_3$

Molar mass of $Al$ is $26.98\text{ g/mol}$, $N$ is $14.01\text{ g/mol}$ and $O$ is $16.00\text{ g/mol}$. $M = 26.98\text{ g/mol}+3\times14.01\text{ g/mol}+9\times16.00\text{ g/mol}=26.98\text{ g/mol}+42.03\text{ g/mol}+144.0\text{ g/mol}=213.01\text{ g/mol}$

Step2: Calculate percent composition

Percent of $Al=\frac{26.98\text{ g/mol}}{213.01\text{ g/mol}}\times 100%\approx12.67%$ Percent of $N=\frac{3\times14.01\text{ g/mol}}{213.01\text{ g/mol}}\times 100%\approx19.77%$ Percent of $O=\frac{9\times16.00\text{ g/mol}}{213.01\text{ g/mol}}\times 100%\approx67.56%$

Answer:

$Al: 12.67%, N: 19.77%, O: 67.56%$

b. $NH_4Cl$

Explanation:

Step1: Find molar mass of $NH_4Cl$

Molar mass of $N$ is $14.01\text{ g/mol}$, $H$ is $1.01\text{ g/mol}$, $Cl$ is $35.45\text{ g/mol}$. $M = 14.01\text{ g/mol}+4\times1.01\text{ g/mol}+35.45\text{ g/mol}=14.01\text{ g/mol}+4.04\text{ g/mol}+35.45\text{ g/mol}=53.5\text{ g/mol}$

Step2: Calculate percent composition

Percent of $N=\frac{14.01\text{ g/mol}}{53.5\text{ g/mol}}\times 100%\approx26.2%$ Percent of $H=\frac{4\times1.01\text{ g/mol}}{53.5\text{ g/mol}}\times 100%\approx7.59%$ Percent of $Cl=\frac{35.45\text{ g/mol}}{53.5\text{ g/mol}}\times 100%\approx66.3%$

Answer:

$N: 26.2%, H: 7.59%, Cl: 66.3%$

9. Determine the empirical formula of the compounds below

a. $69.58%$ Ba, $24.32%$ O, $6.09%$ C

Explanation:

Step1: Assume 100 g of the compound

So we have 69.58 g Ba, 24.32 g O and 6.09 g C.

Step2: Calculate moles of each element

$n_{Ba}=\frac{69.58\text{ g}}{137.33\text{ g/mol}}\approx0.507\text{ mol}$ $n_O=\frac{24.32\text{ g}}{16.00\text{ g/mol}} = 1.52\text{ mol}$ $n_C=\frac{6.09\text{ g}}{12.01\text{ g/mol}}\approx0.507\text{ mol}$

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (0.507 mol). $Ba:O:C = 1:3:1$, empirical formula is $BaCO_3$

Answer:

$BaCO_3$

b. $35.5%$ C, $4.77%$ H, $8.29%$ N, $13.6%$ Na, $37.9%$ O

Explanation:

Step1: Assume 100 g of the compound

So we have 35.5 g C, 4.77 g H, 8.29 g N, 13.6 g Na and 37.9 g O.

Step2: Calculate moles of each element

$n_C=\frac{35.5\text{ g}}{12.01\text{ g/mol}}\approx2.96\text{ mol}$ $n_H=\frac{4.77\text{ g}}{1.01\text{ g/mol}}\approx4.72\text{ mol}$ $n_N=\frac{8.29\text{ g}}{14.01\text{ g/mol}}\approx0.592\text{ mol}$ $n_{Na}=\frac{13.6\text{ g}}{22.99\text{ g/mol}}\approx0.592\text{ mol}$ $n_O=\frac{37.9\text{ g}}{16.00\text{ g/mol}}\approx2.37\text{ mol}$

Step3: Find the mole - ratio

Divide each number of moles by 0.592 mol. $C:H:N:Na:O\approx5:8:1:1:4$, empirical formula is $C_5H_8NNaO_4$

Answer:

$C_5H_8NNaO_4$