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3 years ago

How many moles are in 456 g of CaCO3

Chemistry

1 answer:

GrogVix [38]3 years ago

3 0

Answer:

4.556 moles

Explanation:

mole=mass/ molar mass

mole= 456/100.08

mole= 4.556

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Answer: The empirical formula for the given compound is $C_{2}H_{5}O$

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The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

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Mass of $H_2O=14.6g$

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Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

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In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 28.6 g of carbon dioxide, $\frac{12}{44}\times 28.6=7.8g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.6 g of water, $\frac{2}{18}\times 14.6=1.6$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (14.60) - (7.8 + 1.6) = 5.2 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{7.8g}{12g/mole}=0.65moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.6g}{1g/mole}=1.6moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{5.2g}{16g/mole}=0.32moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.32 moles.

For Carbon = $\frac{0.65}{0.32}=2.03\approx 2$

For Hydrogen = $\frac{1.6}{0.32}=5$

For Oxygen = $\frac{0.32}{0.32}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 1

Hence, the empirical formula for the given compound is $C_{2}H_{5}O_{1}=C_{2}H_{5}O$

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