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Percentage Composition

A sample of a...

Question

A sample of a pure compound contains $9.81 gm$ of $Z n$ , $1.8 \times 10^{23}$ atoms of chromium and $0.60$ mole of oxygen atoms. What is its simplest formula?

Z n C r_{2} O_{7}

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Z n C r_{2} O_{4}

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Z n C r O_{4}

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Z n C r O_{6}

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Solution

The correct option is B $Z n C r_{2} O_{4}$
We know ,
$9.81 g m$ of $Z n = \frac{9.81}{65.3} \approx 0.15 m o l$
$1.8 \times 10^{23} a t o m s o f C r = \frac{1.8 \times 10^{23}}{6.02 \times 10^{23}} \approx 0.3 m o l$
$M o l e o f O a t o m s = 0.60$
So from mole ratio
$Z n : C r : O = 0.15 : 0.3 : 0.60 = 1 : 2 : 4$
So the simplest formula is $Z n C r_{2} O_{4}$

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9.81 gm

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