Chapter 8.3, Problem 37PP

Interpretation Introduction

Interpretation: The mass of sulfuric acid needs to be determined.

Concept introduction: By multiplying the molar mass of a compound by the number of moles, the given moles get converted to mass.

Answer to Problem 37PP

  $\text{318}{\text{.5 grams H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Explanation of Solution

Initially, the molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated.

One mole of hydrogen atoms weighs $1$ gram

But ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has $2$ atoms of H.

  $\therefore$ Mass of one mole of a hydrogen molecule $\left({\text{H}}_{\text{2}}\right)$

  $=2\times 1=2\text{ grams}$

One mole of sulfur (S) atom weighs $\text{32}\cdot \text{0 grams}$

One mole of oxygen (O) atom weighs $\text{16 grams}$

But ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has $4$ atoms of O

  $\therefore$ Mass of $4$ mole of oxygen $=4\times 16\text{ grams = 64 grams}$

Now the molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is:

  $2\text{ grams + 32 grams + 64 grams = 98 grams}$

Hence one mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ molecule weighs $98\text{ grams}$

Given $3\cdot 25\text{ moles}$ of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Multiplying the given moles by molar mass gives the mass

  $3\cdot 25\text{ moles }\times \frac{98{\text{ grams H}}_{\text{2}}{\text{SO}}_{\text{4}}}{1\text{ mole}}=318\cdot 5{\text{ gm H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Conclusion

Moles to mass is converted by multiplying the given moles by molar mass.