Question

Calculate the relative molecular masses of :

(a) $CHC{l}_3$ (b) $(N{H}_4{)}_{2}C{r}_2{O}_7$

(c) $CuS{O}_4.5H_{2}O$ (d) $(N{H}_4{)}_{2}S{O}_4$

(e) $C{H}_{3}COONa$ (f) Potassium chlorate

(g) Ammonium chloroplatinate $(N{H}_4{)}_{2}PtC{l}_6$

Answer 1

(a) Atomic mass of C = 12 u

Atomic mass of H = 1 u

Atomic mass of Cl = 35.5 u

Therefore, molecular mass of $CHC{l}_3$ can be calculated as follows:

$1\times 12$ + $1\times 1$ + $3\times 35.5$ = 119.5

(b) $14.0067+1.00794\times 4$ $\times 2$ + $51.9961\times 2$ + $15.9994\times 7$ = 252.06492 g/mol

(c) Atomic mass of H = 1 g
atomic mass of O = 16 g
atomic mass of Cu = 63.5 g
atomic mass of S = 32 g

now,
molar mass of $CuS{O}_4.5H_{2}O$ = atomic mass of Cu + atom mass of S + $4\times$ atomic mass of O + $5\times 2\times atomicmassofH+atomicmassofO$

= $63.5+32+4\times 16+5$ $2+16$ g

= $63.5+32+64+90$ g

= 249.5 g

(d) The molecular mass of $(N{H}_4{)}_{2}S{O}_4$
= $2\times (14+4\times 1)$ +$32+4\times 16$
= 132 g/ mole

(e) $C{H}_{3}COONa$ :
C = 12
H = 1
O = 16
Na = 23

= $12+1\times 3+12+16\times 2+23$
= $12+3+12+32+23$
= 82g

(f) K=39.1 g /mol

Cl = 35.5 g/mol

O = 16.0 g/mol *3

$39.1+35.5+(16.0\times 3)$ = 122.6 g/mol

(g) Ammonium hexachloroplatinate, also known as ammonium chloroplatinate, is the inorganic compound with the formula $\left(NH4{)}_2\right[PtC{l}_6]$. It is a rare example of a soluble platinum(IV) salt that is not hygroscopic. It forms intensely yellow solutions in water. In the presence of 1M $N{H}_{4}Cl$ , its solubility is only 0.0028 g/100 mL.