Question #ed77e

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Question #ed77e

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Answer 1 · 2017-04-13T01:14:09

$C_{1} H_{2} O_{1}$ $C_1H_2O_1$

Explanation:

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$1) Assume 100 g sample and multiply by percentages to find mass of each element$ $color(blue)("1) Assume 100 g sample and multiply by percentages to find mass of each element")$

$C = 39.99 g$ $C = 39.99g$
$H = 6.79 g$ $H = 6.79g$
$O = 53.28 g$ $O = 53.28g$

$2)Take mass of each element and convert to moles using molar mass from periodic table$ $color(blue)("2)Take mass of each element and convert to moles using molar mass from periodic table")$

$Carbon \to \frac{39.99 g}{1} \cdot \frac{1 mol}{12 g} = 3.33 mol C$ $"Carbon"->(39.99g)/(1)*(1 "mol")/(12g) = 3.33" mol C"$

$Hydrogen \to \frac{6.79 g}{1} \cdot \frac{1 mol}{1.00 g} = 6.79 mol H$ $"Hydrogen"->(6.79g)/(1)*(1 "mol")/(1.00g) = 6.79" mol H"$

$Oxygen \to \frac{53.28 g}{1} \cdot \frac{1 mol}{16 g} = 3.33 mol O$ $"Oxygen"->(53.28g)/(1)*(1 "mol")/(16g) = 3.33" mol O"$

$3)Divide the moles by the smallest mole number of the three values$ $color(blue)("3)Divide the moles by the smallest mole number of the three values")$

$Carbon \to \frac{3.33}{3.33} = 1$ $"Carbon"-> (3.33)/(3.33) = 1$
$Hydrogen \to \frac{6.79}{3.33} = 2.03 \approx 2$ $"Hydrogen"->(6.79)/(3.33) = 2.03 approx 2$
$Oxygen \to \frac{3.33}{3.33} = 1$ $"Oxygen"->(3.33)/(3.33) = 1$

$4)Write out the element with their respected whole number mole ratios$ $color(blue)("4)Write out the element with their respected whole number mole ratios")$

$C_{1} H_{2} O_{1}$ $C_1H_2O_1$

Answer: $C_{1} H_{2} O_{1}$ $C_1H_2O_1$

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Question #ed77e

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