The individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.
Moles of CO2 formed can be calculated as
= Mass of CO2 / Molar mass of CO2
= 1.9061 / 44 = 0.0433 moles
<h3>Calculation of no. of moles of carbon</h3>
Now, moles of C which is present in one mole of CO2 = 1 mole
Moles of C in 0.0433 moles of CO2 = 0.0433 moles
As we know that, molar mass of C = 12 g / mol
Mass of C in 0.7549 g of given sample can be calculated as
= 0.0433 × 12 =0.5196 g
Mass of H2O formed = 0.3370 g
Similarly, Molar Mass of H2O = 18 g / mol
Moles of H2O = 0.3370 / 18 = 0.0187 moles
Moles of H present in 1 mole of H2O = 2 moles
Moles of H present in 0.0187 mole of H2O = 2 × 0.0187 = 0.0374 moles
Molar mass of H = 1 g / mol
Mass of H contained in 0.7549 g of sample = 1 × 0.0374= 0.0374 g
Mass of O in 0.7549 g sample can be calculated as
= 0.7549 – [(Mass of C ) + (Mass of H) ]
= 0.7549 – [ (0.5196) + (0.0374) ]
= 0.1979 g
Thus, we calculated that the individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.
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DISCLAIMER: THE above question is incomplete. Complete question is given below:
A 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g). Calculate the individual mass
of C, H and O in the given sample.
= 3.2 X 