Question

A 0.67 gram sample of chromium is reacted with sulfur. The resulting chromium sulfide has a mass of 1.2888 grams. What is the empirical formula

Answer 1

Answer:

Cr₂S₃

Explanation:

From the question given above, the following data were obtained:

Mass of chromium (Cr) = 0.67 g

Mass of chromium sulfide = 1.2888 g

Empirical formula =?

Next, we shall determine the mass of sulphur (S) in the compound. This can be obtained as follow:

Mass of chromium (Cr) = 0.67 g

Mass of chromium sulfide = 1.2888 g

Mass of sulphur (S) =?

Mass of S = (Mass of chromium sulfide) – (Mass of Cr)

Mass of S = 1.2888 – 0.67

Mass of S = 0.6188 g

Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:

Mass of Cr = 0.67 g

Mass of S = 0.6188 g

Divide by their molar mass

Cr = 0.67 / 52 = 0.013

S = 0.6188 / 32 = 0.019

Divide by the smallest

Cr = 0.013 / 0.013 = 1

S = 0.019 / 0.013 = 1.46

Multiply by 2 to express in whole number

Cr = 1 × 2 = 2

S = 1.46 × 2 = 3

Therefore, the empirical formula of the compound is Cr₂S₃

Answer 2

The empirical formula will be "$Cr_2 S_3$".

Given:

Mass of Cr,

• 0.67 gram

Mass of product,

• 1.2888 grams

Now,

→ $Mass \ of \ Sulphur = Mass \ of \ product -Mass \ of \ Cr$

                                $= 1.2888-0.67$

                                $= 0.6188 \ g$

then,

→ $Moles \ ratio \ of \ Cr:S = (\frac{0.67}{51.996} ): (\frac{0.61888}{32} )$

                                       $= 0.0129: 0.0194$

                                       $= 2:3$

Thus the above response is right.

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