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3 months ago

What is the empirical formula of a compound that contains 27.0% s, 13.4% o, and 59.6% cl by mass?

Chemistry

2 answers:

alisha [2.9K]3 months ago

7 0

Answer: The empirical formula for a compound is $SOCl_2$

Solution:

When percentages are provided, we assume a total mass of 100 grams.

This means that the mass of each element corresponds to its given percentage.

Mass of S = 27.0 g

Mass of O = 13.4 g

Mass of Cl = 59.6 g

Molar mass of S = 32 g/mole

Molar mass of O = 16 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1: Convert the given masses to moles.

Moles of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{27.0g}{32g/mole}=0.844moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{13.4g}{16g/mole}=0.837moles$

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{59.6g}{35.5g/mole}=1.68moles$

Step 2: For the mole ratio, divide each mole amount by the smallest value found.

For S = $\frac{0.844}{0.837}=1.00\aprrox 1$

For O = $\frac{0.837}{0.837}=1$

For Cl = $\frac{1.68}{0.837}=2.00\approx 2$

The ratio of S: O: Cl = 1: 1: 2

We represent this mole ratio as subscripts in the empirical formula.

The empirical formula = $S_1O_1Cl_2=SOCl_2$

Thus, the empirical formula for the compound is $SOCl_2$

castortr0y [3K]3 months ago

6 0

27g of sulfur divided by 32.065g per mole gives 0.84 moles.
13.4g of oxygen divided by 16g per mole gives 0.83 moles.
Take each of these results, divide by the smallest quantity, and round to the nearest whole number.

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