Chapter 6, Problem 51A

Interpretation Introduction

Interpretation: Using the atomic mass or molar mass from periodic table molar mass in column 2 must be matched with the substances in column 1.

Concept Introduction :

Atomic mass is the “average” of all atoms for an element. Molar mass is the total mass of all atoms in 1 mol of compound.

Answer to Problem 51A

1 will match with c.

2 will match with e.

3 will match with j.

4 will match with h.

5 will match with b.

6 will match with d.

7 will match with a.

8 will match with g.

9 will match with i.

10 will match with f.

Explanation of Solution

1 will match with ‘c’ as molybdenum (Mo) has atomic mass of 95.94 g.

2 will match with ‘e’ as atomic mass of lanthanum (La) is 138.9 g.

3 will match with ‘j’ as molar mass of carbon tetrabromide (CBr₄) is $\text{=(12}\text{.01+4}\times \text{79}\text{.9) g=(12}\text{.0+319}\text{.6) g=331}\text{.6 g}\text{.}$

4 will match with ‘h’ as molar mass of mercury (II) oxide (HgO) is $\text{=(200}\text{.6+16) g=216}\text{.6 g}\text{.}$

5 will match with ‘b’ as molar mass of titanium (IV) oxide (TiO₂) $\text{=(47}\text{.88+2}\times 16\text{) g=(47}\text{.88+32) g=79}\text{.88 g}\text{.}$

6 will match with‘d’ as molar mass of manganese (II) chloride (MnCl₂) $\text{=(54}\text{.84+71) g=125}\text{.84 g}\text{.}$

7 will match with ‘a’ as molar mass of phosphine (PH₃) $\text{=(30}\text{.99+3}\times \text{1) g=33}\text{.99 g}\text{.}$

8 will match with ‘g’ as molar mass of tin (II) fluoride (SnF₂) $\text{=(118}\text{.7+2}\times \text{19) g=(118}\text{.7+38) g=156}\text{.7 g}\text{.}$

9 will match with ‘i’ as molar mass of lead (II) sulfide (PbS) $\text{=(207}\text{.3+32) g=239}\text{.3 g}\text{.}$

10 will match with ‘f’ as molar mass of copper (I) oxide (Cu₂O) $\text{=(2}\times \text{63}\text{.55+16) g=(127}\text{.1+16) g=143}\text{.1 g}\text{.}$

Conclusion

The atomic or molar mass in column II is matched with column I.