Question

1. how many molecules are in each sample?

a. 6.5 g h₂o
b. 389 g cbr₄
c. 22.1 g o₂
d. 19.3 g c₈h₁₀

Explanation:

Step1: Recall the molar - mass formula and Avogadro's number

First, find the molar mass ($M$) of each compound using the atomic masses of the elements. Then use the formula $n=\frac{m}{M}$ (where $n$ is the number of moles, $m$ is the mass in grams) and the fact that 1 mole of any substance contains $N_A = 6.022\times10^{23}$ molecules.

Step2: Calculate molar mass of $H_2O$

The atomic mass of $H$ is approximately $1.01\ g/mol$ and of $O$ is approximately $16.00\ g/mol$. So, $M_{H_2O}=2\times1.01 + 16.00=18.02\ g/mol$.
$n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{6.5\ g}{18.02\ g/mol}\approx0.361\ mol$.
The number of molecules of $H_2O$ is $N_{H_2O}=n_{H_2O}\times N_A=0.361\ mol\times6.022\times 10^{23}\ molecules/mol\approx2.17\times 10^{23}$ molecules.

Step3: Calculate molar mass of $CBr_4$

The atomic mass of $C$ is approximately $12.01\ g/mol$ and of $Br$ is approximately $79.90\ g/mol$. So, $M_{CBr_4}=12.01+4\times79.90 = 12.01 + 319.6=331.61\ g/mol$.
$n_{CBr_4}=\frac{m_{CBr_4}}{M_{CBr_4}}=\frac{389\ g}{331.61\ g/mol}\approx1.17\ mol$.
The number of molecules of $CBr_4$ is $N_{CBr_4}=n_{CBr_4}\times N_A=1.17\ mol\times6.022\times 10^{23}\ molecules/mol\approx7.04\times 10^{23}$ molecules.

Step4: Calculate molar mass of $O_2$

The atomic mass of $O$ is approximately $16.00\ g/mol$. So, $M_{O_2}=2\times16.00 = 32.00\ g/mol$.
$n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{22.1\ g}{32.00\ g/mol}\approx0.691\ mol$.
The number of molecules of $O_2$ is $N_{O_2}=n_{O_2}\times N_A=0.691\ mol\times6.022\times 10^{23}\ molecules/mol\approx4.16\times 10^{23}$ molecules.

Step5: Calculate molar mass of $C_8H_{10}$

The atomic mass of $C$ is approximately $12.01\ g/mol$ and of $H$ is approximately $1.01\ g/mol$. So, $M_{C_8H_{10}}=8\times12.01+10\times1.01=96.08 + 10.1=106.18\ g/mol$.
$n_{C_8H_{10}}=\frac{m_{C_8H_{10}}}{M_{C_8H_{10}}}=\frac{19.3\ g}{106.18\ g/mol}\approx0.182\ mol$.
The number of molecules of $C_8H_{10}$ is $N_{C_8H_{10}}=n_{C_8H_{10}}\times N_A=0.182\ mol\times6.022\times 10^{23}\ molecules/mol\approx1.10\times 10^{23}$ molecules.

Answer:

a. Approximately $2.17\times 10^{23}$ molecules
b. Approximately $7.04\times 10^{23}$ molecules
c. Approximately $4.16\times 10^{23}$ molecules
d. Approximately $1.10\times 10^{23}$ molecules