Question

(a) A compound contains 47% = K 14.5%=C, 38.5%=0. Calculate empirical formula and
molecular formula. (Atomic mass of C = 12,0 = 16, K = 39]​

Answer 1

K=47percantage c=14.5,O=38.5

Convert into moles

47/39 =1.20

14.5/12=1.20

38.5/16=1.40

1.20, 1.20, 1.40 take ratio we get 1:1:2 = KCO2= empirical formula

Molecular formula =(empirical formula )n

Answer 2

Empirical formula and  molecular formula are $KCO_2$  and $K_2C_2O_4$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of K= 47 g

Mass of C = 14.5 g

Mass of O = 38.5 g

Step 1 : convert given masses into moles

Moles of K =$\frac{\text{ given mass of K}}{\text{ molar mass of K}}= \frac{47g}{39g/mole}=1.20moles$

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{14.5g}{12g/mole}=1.20moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.5g}{16g/mole}=2.40moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For K = $\frac{1.20}{1.20}=1$

For C = $\frac{1.20}{1.20}=1$

For O =$\frac{2.40}{1.20}=2$

The ratio of K : C: O= 1: 1: 2

Hence the empirical formula is $KCO_2$

The empirical weight of = 1(39)+1(12)+2(16)= 83 g.

The molecular weight = 166.22 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{166.22}{83}=2$

The molecular formula will be=$2\times KCO_2=K_2C_2O_4$

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