Chapter 18, Problem 119A

(a)

Interpretation Introduction

Interpretation: The mass of $\text{4}\text{.50 mol Fe}$ is to be calculated.

Concept Introduction: The mass is directly proportional to the number of moles and inversely proportional to the molar mass.

Answer to Problem 119A

The mass of $\text{4}\text{.50 mol Fe}$ is $251.3025\text{ g}$ .

Explanation of Solution

The number of moles of $\text{Fe}$ (iron) is $\text{4}\text{.50 mol}$ .

The molar mass of $\text{Fe}$ is $\text{55}{\text{.845 g mol}}^{-1}$ .

The formula to calculate mass is as follows:

  $m=nM$

Where,

• $m$ is the mass.
• $M$ is the molar mass.
• $n$ is the number of moles.

Substitute the values of $n$ and $M$ in the above formula.

  $\begin{array}{c}m=nM\\ =\text{4}\text{.50 }\bcancel{\text{mol}}\times \text{55}\text{.845 g }\bcancel{{\text{mol}}^{-1}}\\ =251.3025\text{ g}\end{array}$

Therefore, the mass of $\text{4}\text{.50 mol Fe}$ is $251.3025\text{ g}$ .

(b)

Interpretation Introduction

Interpretation: The mass in grams of $\text{36}\text{.8 L}$

  $\text{CO}$ (at STP) is to be calculated.

Concept Introduction: Abbreviation Standard Temperature and Pressure is abbreviated as "STP." At STP, one mole of any gas will occupy a volume of $22.4\text{ L}$ .

Answer to Problem 119A

The mass of $\text{36}\text{.8 L}$

  $\text{CO}$ (at STP) is $46\text{ g}$ .

Explanation of Solution

The volume of $\text{CO}$ (carbon dioxide) at STP is $\text{36}\text{.8 L}$ .

One mole of $\text{CO}$ is occupied $22.4\text{ L}$ .

The molar mass of $\text{CO}$ is ${\text{28 g mol}}^{-1}$ .

The formula to calculate the mass of $\text{CO}$

(at STP) is as follows:

  $m=\frac{1\text{ mol}}{22.4\text{ L}}VM$

Where,

• $m$ is the mass.
• $M$ is the molar mass.
• $V$ is the volume.

Substitute the values of $V$ and $M$ in the above formula.

  $\begin{array}{c}m=\frac{1\text{ mol}}{22.4\text{ L}}VM\\ =\frac{\text{1 }\bcancel{\text{mol}}}{22.4\bcancel{\text{L}}}\times \text{36}\text{.8 }\bcancel{\text{L}}\times \text{28 g }\bcancel{{\text{mol}}^{-1}}\\ =46\text{ g}\end{array}$

Therefore, the mass of $\text{36}\text{.8 L}$

  $\text{CO}$ (at STP) is $46\text{ g}$ .

(c)

Interpretation Introduction

Interpretation: The mass in grams of $1\text{ molecule}$ of glucose is to be calculated.

Concept Introduction: One mole of any substance contains $6.022\times {10}^{23}$ particles.

Answer to Problem 119A

The mass of $1\text{ molecule}$ of glucose is $2.9890\times {10}^{-22}\text{ g}$ .

Explanation of Solution

The number of molecules of glucose is $1\text{ molecule}$ .

The molar mass of glucose is ${\text{180 g mol}}^{-1}$ .

The formula to calculate the mass of glucose is as follows:

  $m=\frac{1\text{ mol}\times 1\text{ molecule}}{6.022\times {10}^{23}\text{molecule}}M$

Where,

• $m$ is the mass.
• $M$ is the molar mass.

Substitute the value of $M$ in the above formula.

  $\begin{array}{c}m=\frac{1\text{ mol}\times 1\text{ molecule}}{6.022\times {10}^{23}\text{molecule}}M\\ =\frac{1\bcancel{\text{mol}}\times 1\bcancel{\text{molecule}}}{6.022\times {10}^{23}\bcancel{\text{molecule}}}\times 180\text{ g }\bcancel{{\text{mol}}^{-1}}\\ =2.9890\times {10}^{-22}\text{ g}\end{array}$

Therefore, the mass of $1\text{ molecule}$ of glucose is $2.9890\times {10}^{-22}\text{ g}$ .

(d)

Interpretation Introduction

Interpretation: The mass of $\text{0}\text{.0642 mol}$ ammonium phosphate is to be calculated.

Concept Introduction: The mass is directly proportional to the number of moles and inversely proportional to the molar mass.

Answer to Problem 119A

The mass of $\text{0}\text{.0642 mol}$ ammonium phosphate is $9.5658\text{ g}$ .

Explanation of Solution

The number of moles of ammonium phosphate is $\text{0}\text{.0642 mol}$ .

The molar mass of ammonium phosphate is ${\text{149 g mol}}^{-1}$ .

The formula to calculate mass is as follows:

  $m=nM$

Where,

• $m$ is the mass.
• $M$ is the molar mass.
• $n$ is the number of moles.

Substitute the values of $n$ and $M$ in the above formula.

  $\begin{array}{c}m=nM\\ =\text{0}\text{.0642 }\bcancel{\text{mol}}\times \text{149 g }\bcancel{{\text{mol}}^{-1}}\\ =9.5658\text{ g}\end{array}$

Therefore, the mass of $\text{0}\text{.0642 mol}$ ammonium phosphate is $9.5658\text{ g}$ .