Question

# combustion of 1.00 g of a common analgesic which is composed of carbon, hydrogen, oxygen produced...

combustion of 1.00 g of a common analgesic which is composed of carbon, hydrogen, oxygen produced 2.20g of carbon dioxide, CO2 and 0.440 grams of water. The molar mass of this compound is known to be between 170 and 190 g. Determine the molecular formula of the compound.

#### Homework Answers

Answer #1

Solution :-

Lets first calculate the mass of the each element using the given mass of the CO2 , H2O and sample mass

mass of the C = 2.20 g CO2 * 27.29% C / 100 % = 0.60 g C

mass of H = 0.440 g H2O * 11.2 % H / 100 % = 0.0493 g H

now lets find the mass of oxygen

mass of O = mass of sample - (mass of C + mass of H)

= 1.00 g - (0.60 g +0.0493 g)

= 0.351 g

now lets calculate moles of the each element

moles = mass / molar mass

moles of C = 0.600 g / 12.01 g per mol = 0.05 mol

moles of H = 0.0493 g / 1.0079 g per mol = 0.05 mol

moles of O = 0.351g / 16 g per mol = 0.022 mol

now lets find the ratio of the elements by dividing the moles by the smallest mole value

C= 0.05 / 0.022 = 2

H= 0.05 / 0.022 = 2

O = 0.022 / 0.022 = 1

Therefore Empirical formula is C2H2O

now lets find the empirical fiormula mass

C2H2O = (2*12.01)+(2*1.0079)+(1*16) = 42.04 g

now lets find the ratio of the molar mass to empirical formula mass

N= molar mass / empirical formula mass

= 170 / 42.04

= 4

Therefore moilecular formula = 4 * empirical formula

= 4*C2H2O

= C8H8O4

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