combustion analysis of an unknown compound provides the following data: 73.5 grams carbon (C), 4.20 grams hydrogen (H) and 72.3
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Answer: .
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 K = 49.4 g
Mass of S = 20.3 g
Mass of O = 30.3 g
Step 1 : convert given masses into moles.
Moles of K=
Moles of S= \frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{20.3g}{32g/mole}=0.63moles[/tex]
Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{30.3g}{16g/mole}=1.89moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For K =
For S =
For O =
The ratio of K: S:O = 2: 1: 3
Hence the empirical formula is .