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Determine the empirical formulas for compounds with the following percent compositions:
(a)15.8% carbon and 84.2 percent sulfur
(b)40.0 percent carbon, 6.7% hydrogen, and 53.3% oxygen

Respuesta :

Answer: a)  [tex]CS_2[/tex]

b) [tex]CH_2O[/tex]

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

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

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

For C = [tex]\frac{1.32}{1.32}=1[/tex]

For S =[tex]\frac{2.63}{1.32}=2[/tex]

The ratio of C  : S = 1:2

Hence the empirical formula is [tex]CS_2[/tex]

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

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

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

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

For C = [tex]\frac{3.33}{3.33}=1[/tex]

For H = [tex]\frac{6.7}{3.33}=2[/tex]

For O =[tex]\frac{3.33}{3.33}=1[/tex]

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

Hence the empirical formula is [tex]CH_2O[/tex]

Eduard22sly

A. The empirical formula of the compound containing 15.8% carbon and 84.2% sulphur is CS₂

B. The empirical formula of the compound containing 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen is CH₂O

A. How to determine the empirical formula

  • Carbon (C) = 15.8%
  • Sulphur (S) = 84.2%
  • Empirical formula =?

Divide by their molar mass

C = 15.8 / 12 = 1.317

S = 84.2 / 32 = 2.631

Divide by the smallest

C = 1.317 / 1.317 = 1

S = 2.631 / 1.317 = 2

Thus, the empirical formula of the compound is CS₂

B. How to determine the empirical formula

  • Carbon (C) = 40%
  • Hydrogen (H) = 6.7%
  • Oxygen (O) = 53.3%
  • Empirical formula =?

Divide by their molar mass

C = 40 / 12 = 3.33

H = 6.7 / 1 = 6.7

O = 53.3 / 16 = 3.33

Divide by the smallest

C = 3.33 / 3.33 = 1

H = 6.7 / 3.33 = 2

O = 3.33 / 3.33 = 1

Thus, the empirical formula of the compound is CH₂O

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