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2 years ago

What is the empirical formula for a compound that contains 10.89% magnesium 31.77% chloride and 57.34% oxygen

Chemistry

1 answer:

Blizzard [7]2 years ago

4 0

Answer: $Mg_{1}Cl_{2}O_{8}$

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 Mg = 10.89 g

Mass of Cl = 31.77 g

Mass of O = 57.34 g

Step 1 : convert given masses into moles.

Moles of Mg= $\frac {\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac {10.89g}{24g/mole}=0.45moles$

Moles of Cl = $\frac {\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac {31.77g}{35.5g/mole}=0.89moles$

Moles of O = $\frac {\text{ given mass of O}}{\text{ molar mass of O}}=\frac {57.34g}{16g/mole}=3.58moles$

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

For Mg = $\frac {0.45}{0.45}=1$

For Cl = $\frac {0.89}{0.45}=2$

For O= $\frac {3.58}{0.45}=8$

The ratio of Mg :Cl : O= 1 : 2 : 8

Hence the empirical formula is $Mg_{1}Cl_{2}O_{8}$

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