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Answer 1 · 2017-04-13T01:14:09

$C_1H_2O_1$

Let's try to work this through.

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

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

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

$"Carbon"->(39.99g)/(1)*(1 "mol")/(12g) = 3.33" mol C"$

$"Hydrogen"->(6.79g)/(1)*(1 "mol")/(1.00g) = 6.79" mol H"$

$"Oxygen"->(53.28g)/(1)*(1 "mol")/(16g) = 3.33" mol O"$

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

$"Carbon"-> (3.33)/(3.33) = 1$
$"Hydrogen"->(6.79)/(3.33) = 2.03 approx 2$
$"Oxygen"->(3.33)/(3.33) = 1$

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

$C_1H_2O_1$

Answer: $C_1H_2O_1$