Chapter 10, Problem 117A

Interpretation Introduction

Interpretation:

Whether 10.0 g of C or 10.0 g of $\text{Ca}$ contains more atoms should be determined. The number of atoms of 10.0 g of C and 10.0 g of $\text{Ca}$ should be calculated.

Concept introduction:

The number of moles of a compound is defined as the ratio of the given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = $\frac{\text{given mass of the compound}}{\text{molar mass of the compound}}$

The one mole of any substance is equal to the Avogadro number, i.e., $6.022\times {10}^{23}$ atoms.

Answer to Problem 117A

The number of atoms of 10.0 g of carbon is greater the number of atoms of 10.0 g of calcium.

Number of atoms of $\text{C}$ = $\text{5}\text{.014}\times {\text{10}}^{23}\text{ atoms}$

Number of atoms of $\text{Ca}$ = $1.50\times {10}^{23}\text{ atoms}$

Explanation of Solution

Mass of carbon = 10.0 g

Mass of calcium = 10.0 g

Molar mass of carbon = 12.0107 g/mole

Molar mass of calcium = 40.078 g/mole

$\text{Number of moles of carbon = }\frac{\text{10}\text{.0 g}}{12.0107\text{ g/mole}}$

Number of moles of $\text{C}$ = $\text{0}\text{.8326 mole}$

Since one mole of any substance is equal to the Avogadro number, i.e., $6.022\times {10}^{23}$ atoms.

Number of moles = $\text{number of atoms ×}\frac{1\text{ mole}}{6.022\times 10\text{atoms}}$

Put the values,

$\text{0}\text{.8326 mole}$ = $\text{Number of atoms ×}\frac{1\text{ mole}}{6.022\times 10\text{atoms}}$

$\text{Number of atoms }=\text{0}\text{.8326 mole}\times 6.022\times {10}^{23}\frac{\text{atoms}}{\text{mole}}$

Number of atoms of $\text{C}$ = $\text{5}\text{.014}\times {\text{10}}^{23}\text{ atoms}$

$\text{Number of moles of Ca = }\frac{\text{10}\text{.0 g}}{40.078\text{ g/mole}}$

Number of moles of $\text{Ca}$ = $\text{0}\text{.2495 mole}$

Put the values,

$\text{0}\text{.2495 mole}$ = $\text{Number of atoms ×}\frac{1\text{ mole}}{6.022\times 10\text{atoms}}$

$\text{Number of atoms }=\text{0}\text{.2495 mole}\times 6.022\times {10}^{23}\frac{\text{atoms}}{\text{mole}}$

Number of atoms of $\text{Ca}$ = $1.50\times {10}^{23}\text{ atoms}$

The number of atoms of 10.0 g of carbon is greater than the number of atoms of 10.0 g of calcium.