In an experiment, a 2.514 g sample of calcium is heated in oxygen and the compound formed has a mass of 3.518 g. What is the empirical formula for the compound?

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In an experiment, a 2.514 g sample of calcium is heated in oxygen and the compound formed has a mass of 3.518 g. What is the empirical formula for the compound?

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Answer 1 · 2017-01-20T07:13:23

We get an empirical formula of $C a O$ $CaO$ .

Explanation:

We can work out the individual masses of the elements in the binary compound, and we convert this to a molar quantity of each element:

$Moles of calcium$ $"Moles of calcium"$ $=$ $=$ $\frac{2.514 \cdot g}{40.1 \cdot g \cdot m o l^{- 1}} = 0.0627 \cdot m o l$ $(2.514*g)/(40.1*g*mol^-1)=0.0627*mol$

$Moles of oxygen$ $"Moles of oxygen"$ $=$ $=$ $\frac{3.518 \cdot g - 2.514 \cdot g}{16.00 \cdot g \cdot m o l^{- 1}} = 0.0627 \cdot m o l$ $(3.518*g-2.514*g)/(16.00*g*mol^-1)=0.0627*mol$

We divide thru by the smallest molar quantity, and since there are equimolar quantities of metal and oxygen, the empirical formula is:

$C a O$ $CaO$ .

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In an experiment, a 2.514 g sample of calcium is heated in oxygen and the compound formed has a mass of 3.518 g. What is the empirical formula for the compound?

1 Answer

Explanation:

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