An unknown compound contains only carbon, hydrogen, and oxygen (cxhyoz). combustion of 4.50 g of this compound produced 6.60 g o f carbon dioxide and 2.70 g of water. how many moles of hydrogen, h, were in the original sample
1 answer:
The number of moles of Hydrogen H that were in the original sample is
<u>0.30 moles </u>
<h3>calculation</h3><h3> f<u>
ind the formula of the unknown compound </u></h3>
write the equation for reaction CxHyOz + O2 → CO2 + H2O
find the moles of CO2 produced moles= mass/molar mass
= 6.60g/44 g/mol=0.15 moles
moles of CO2= moles of CxHyOz therefore the moles of CxHyOz =0.15 moles find the molar mass of CxHyOz molar mass= mass/ moles
= 4.50/0.15= 30 g/mol
determine value the for X,Y,Z C =x(12 g/mol) if x=1 then c= 12 g/mol
H= y (1 g/mol) if Y=1 then H=1 g/mol
O=Z(16g/mol) if O=16 then O= 16g/mol
the formula of the compound from the value of X,y and z above = CHO
the MW of CHO= (12 g/mol + 1 g/mol +16g/mol)= 29 g/mol since the molar mass of the compound was 30 g/mol 1 more atom of oxygen is required, therefore the formula of unknown compound= CH2O from the moles of CH2O= 0.15 moles and there are two atoms of H in CH2O therefore the moles of H atom = 0.15 x2=<u>0.30 moles </u> <u> </u>
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