A $15.0 g$ sample of copper carbonate is heated to form $9.65 g$ of $C u O$ and $5.34 g C O_{2}$ .
What is the empirical formula of this copper carbonate sample?

C u_{2} C O_{3}

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C u C O_{3}

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C u C O

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C u_{3} C O_{3}

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Solution

The correct option is B $C u C O_{3}$
copper carbonate $\to C u O + C O_{2}$
$\frac{15}{x g m}$ $\frac{9.65}{79.54} = 0.12$
$\frac{5.34}{44} = 0.12$
copper carbonate is o.12 mol
so $\frac{15}{x}$ = 0.12
the x is $\frac{15}{0.12} = 125$
B is right option

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A 15.0 g sample of copper carbonate is heated to form 9.65 g of CuO and 5.34 g CO2.What is the empirical formula of this copper carbonate sample?

The correct option is B CuCO3copper carbonate →CuO+CO2 15xgm 9.6579.54=0.12 5.3444=0.12copper carbonate is o.12 mol so 15x = 0.12 the x is 150.12=125 B is right option

A $15.0 g$ sample of copper carbonate is heated to form $9.65 g$ of $C u O$ and $5.34 g C O_{2}$ .
What is the empirical formula of this copper carbonate sample?

The correct option is B $C u C O_{3}$
copper carbonate $\to C u O + C O_{2}$
$\frac{15}{x g m}$ $\frac{9.65}{79.54} = 0.12$
$\frac{5.34}{44} = 0.12$
copper carbonate is o.12 mol
so $\frac{15}{x}$ = 0.12
the x is $\frac{15}{0.12} = 125$
B is right option