A sample of a compound contains 7.89 g potassium, 2.42 g carbon, and 9.69 g oxygen. What is the empirical and molecular formulas of this compound, which has a molar mass of 198.22 g/mol?

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Answer 1 · 2017-12-22T11:04:55

The empirical formula is $K C O_{3}$ $KCO_3$ . The molecular formula is $K_{2} C_{2} O_{6}$ $K_2C_2O_6$

Let the molecular formula be $K_{m} C_{n} O_{p}$ $K_mC_nO_p$

The weight of the sample is

$m = 7.89 + 2.42 + 9.69 = 20 g$ $m=7.89+2.42+9.69=20g$

The number of moles of each element is

Potassium $= \frac{7.89}{39.1} = 0.2$ $=7.89/39.1=0.2$

Carbon $= \frac{2.42}{12} = 0.2$ $=2.42/12=0.2$

Oxygen $= \frac{9.69}{16} = 0.6$ $=9.69/16=0.6$

Divide by $0.2$ $0.2$

Potassium $= \frac{0.2}{0.2} = 1$ $=0.2/0.2=1$

Carbon $= \frac{0.2}{0.2} = 1$ $=0.2/0.2=1$

Oxygen $= \frac{0.6}{0.2} = 3$ $=0.6/0.2=3$

Therefore,

The empirial formula is $K C O_{3}$ $KCO_3$

Since the molar mass is $= 198.22$ $=198.22$

and

$M_{K} + M_{C} + M_{O} = 39 + 12 + 3 \cdot 16 = 99$ $M_K+M_C+M_O=39+12+3*16=99$

The molecular formula is $K_{2} C_{2} O_{6}$ $K_2C_2O_6$

And the name of this compound is probably

Potassium peroxydicarbonate