1. Calculate the percent composition for each element in CH 206 Glucose.

%C = 40

%H = 6.67

%O = 53.33

Given

Glucose  C₆H₁₂O₆

Required

The percent composition

Solution

MW Gulucose = 180 g/mol

Ar C = 12 g/mol

Ar H = 1 g/mol

Ar O = 16 g/mol

%C = 6.12/180 x 100% = 40%

%H = 12.1/180 x 100% = 6.67%

%O = 6.16/180 x 100% = 53.33%

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.