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Flauer [41]
3 years ago
9

Determine the empirical formulas for compounds with the following percent compositions:

Chemistry
1 answer:
White raven [17]3 years ago
5 0

Answer: a)  CS_2

b) CH_2O

Explanation:

a) 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 C = 15.8g

Mass of S= 84.2 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{15.8g}{12g/mole}=1.32moles

Moles of S=\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{84.2g}{32g/mole}=2.63moles

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

For C = \frac{1.32}{1.32}=1

For S =\frac{2.63}{1.32}=2

The ratio of C  : S = 1:2

Hence the empirical formula is CS_2

b) Mass of C= 40 g

Mass of H= 6.7 g

Mass of O = 53.3 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40g}{12g/mole}=3.33moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.7g}{1g/mole}=6.7moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles

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

For C = \frac{3.33}{3.33}=1

For H = \frac{6.7}{3.33}=2

For O =\frac{3.33}{3.33}=1

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

Hence the empirical formula is CH_2O

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