Chapter 2.11, Problem 2.16PSP

(a)

Interpretation Introduction

Interpretation:

The number of $\text{O}$ atoms in a $\text{12}\text{.0}\text{g}$ sample of compounds: aspirin, ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$ and ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ has to be calculated.

Explanation of Solution

Aspirin:

The molecular formula of aspirin is ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$.

Mass of the given compound is $\text{12}\text{.0}\text{g}$.

Molar mass of carbon is $\text{12}\text{.011}\text{g/mol}$.

Molar mass of hydrogen is $\text{1}\text{.0079}\text{g/mol}$.

Molar mass of oxygen is $\text{15}\text{.9994}\text{g/mol}$.

Molar mass of the compound is:

    $\text{=}\text{(9×12}\text{.011}\text{g/mol}\text{C)}\text{+}\text{(8}\text{×1}\text{.0079}\text{g/mol}\text{H)}\text{+}\text{(4×15}\text{.994}\text{g/mol}\text{O)}\text{=}\text{180}\text{.1598}\text{g/mol}$.

Amount of aspirin $\text{=}\text{12}\text{.0}\text{g}\text{aspirin}\text{×}\frac{\text{1}\text{mol}\text{aspirin}}{\text{180}\text{.1598}\text{g}}\text{=}\text{0}\text{.0666}\text{mol}\text{aspirin}$.

Number of molecules $\text{=}\text{0}\text{.0666}\text{mol}\text{×}\frac{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}}\text{=}\text{4}{\text{.011×10}}^{\text{22}}\text{molecules}$.

Number of oxygen atoms = $\text{4}\text{.011}\text{×}{\text{10}}^{\text{22}}\text{molecules}\text{×}\frac{\text{4}\text{O}\text{atoms}}{\text{1}\text{molecule}}\text{=}\text{1}\text{.6044}\text{×}{\text{10}}^{\text{23}}\text{O}\text{atoms}$.

In ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$:

Mass of the given compound is $\text{12}\text{.0}\text{g}$.

Molar mass of calcium is $40.078\text{g/mol}$.

Molar mass of phosphorous is $30.9737\text{g/mol}$.

Molar mass of oxygen is $\text{15}\text{.9994}\text{g/mol}$.

Molar mass of the compound is:

$\text{=}\text{(3×40}\text{.078}\text{g/mol}\text{Ca)}\text{+}\text{(2}\text{×30}\text{.9737}\text{g/mol}P\text{)}\text{+}\text{(8×15}\text{.994}\text{g/mol}\text{O)}\text{=}310.1766\text{g/mol}$.

Amount of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ $\text{=}\text{12}\text{.0}\text{g}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}{310.1766\text{g}}\text{=}\text{0}\text{.0386}\text{mol}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$.

Number of molecules $\text{=}\text{0}\text{.0386}\text{mol}\text{×}\frac{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}}\text{=}2.3297\times {10}^{22}\text{molecules}$.

Number of oxygen atoms $2.3297\text{×}{\text{10}}^{\text{22}}\text{molecules}\text{×}\frac{8\text{O}\text{atoms}}{\text{1}\text{molecule}}\text{=}\text{1}\text{.8638}\text{×}{\text{10}}^{\text{23}}\text{O}\text{atoms}$.

(b)

Interpretation Introduction

Interpretation:

The mass of compounds ${\text{Mn}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ and ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ that contains $\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{O}$ atoms has to be calculated.

Explanation of Solution

${\text{Mn}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$:

The number of $\text{O}$ atoms is $\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}$.

Molar mass of $\text{Mn}$ is $\text{54}\text{.9380}\text{g/mol}$.

Molar mass of $\text{S}$ is $\text{32}\text{.065}\text{g/mol}$.

Molar mass of $\text{O}$ is $15.9994\text{g/mol}$.

Molar mass of the compound is:

$\text{=}\text{(2×}\text{54}\text{.9380}\text{g/mol}\text{Mn)}\text{+}\text{(3×32}\text{.065}\text{g/mol}\text{S)}\text{+}\text{(12×15}\text{.9994g/mol}\text{O)}\text{=}\text{397}\text{.99}\text{g/mol}$.

Number of molecules $\text{=}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{O}\text{atoms}}{\text{12}\text{O}\text{atoms}}\text{×}\text{1}\text{mol}\text{=}\text{5}{\text{.018×10}}^{\text{22}}\text{molecules}$.

Amount of ${\text{Mn}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ $\text{=}\frac{\text{5}{\text{.018×10}}^{\text{22}}\text{molecules}}{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}\text{×}\text{1}\text{mol}\text{=}\text{.0833}\text{mol}$.

Mass of the compound is $\text{0}\text{.0833}\text{mol}\text{×}\text{397}\text{.99}\text{g}{\text{Mn(SO}}_{\text{4}}{\text{)}}_{\text{3}}\text{=}\text{33}\text{.165}\text{g}$.

${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$:

The number of $\text{O}$ atoms is $\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}$.

Molar mass of $\text{O}$ is $15.9994\text{g/mol}$.

Molar mass of carbon is $\text{12}\text{.011}\text{g/mol}$.

Molar mass of hydrogen is $\text{1}\text{.0079}\text{g/mol}$.

Molar mass of nitrogen is $\text{14}\text{.0067}\text{g/mol}$.

Molar mass of the compound is:

$\text{=}\text{(8×}\text{12}\text{.011}\text{g/mol}\text{C)}\text{+}\text{(10×1}\text{.0079}\text{g/mol}\text{H)}\text{+}\text{(4×14}\text{.0067g/mol}\text{N)}\text{+}\text{(2}\text{×}\text{15}\text{.994g/mol}\text{O)}\text{=}\text{194}\text{.1926g/mol}$.

Number of molecules $\text{=}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{O}\text{atoms}}{\text{2}\text{O}\text{atoms}}\text{×}\text{1}\text{mol}\text{=}3.011{\text{×10}}^{\text{23}}\text{molecules}$.

Amount of ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ $\text{=}\frac{\text{3}{\text{.011×10}}^{\text{23}}\text{molecules}}{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}\text{×}\text{1}\text{mol}\text{=}0.\text{5}\text{mol}$.

Mass of the compound is $\text{0}\text{.5}\text{mol}\text{×}\text{194}\text{.1926}\text{g}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\text{=}\text{9}\text{.74}\text{g}$.