An unknown compound contains only C, H, and O. Combustion of 2.90 g of this compound produced 5.79 g of CO2 and 2.37 g of H2O.What is the empirical formula of the unknown compound?

Md Ariful Islam

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Atomic mass of H=1 g/mole Atomic mass of C=12 g/mole Atomic mass of O=16 g/mole Molar mass of CO2=44 g/mole Molar mass of H2O=18 g/mole Mass of C in 5.79g of CO2 = 5.79/44 g/mole*12 g/mole =1,5792 g Mass of H in 2.37g of H2O=2.37/18 g/mole*2 g/mole =0,2634 g These two amounts of C and H have been obtained from 2.90 g compound containing C ,H and O. So the remaining (2.90-1,5792-0,2634)g=1,0574g is the mass of Oxygen present in 2.9 g compound. Then we must dividing these masses of elements with respective atomic masses of elements and taking their ratio. Then we get the ratio of number of atoms of C, H and O in the molecule of the compound as below C:H:O=(1,5792/12): (0,2634/1): (1,0574/16)= 0,1316:0,2634:0,0660875=(0,1316/0,0660875): (0,2634/0,0660875): (0,0660875/0,0660875)=2:4:1
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If 85.0 L of natural gas, which is essentially methane (CH4), undergoes complete combustion at STP, how many Liters of CO2 is produced CH4 + O2 → CO2+ H2O

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C + 2 H2O  CH4 + CO2 C + H2O  CO + H2 ∆H = +131.3 kJ CO + H2O  CO2 + H2 ∆H = -41.2 kJ CH4 + H2O  3 H2 + CO ∆H = +206.1 kJ

131.3+-41.2-206.1= 116kJ

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Complete combustion of 8.10 g of a hydrocarbon produced 26.0 g of CO2 and 8.88 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary.

CxHy + O2= xCO2 + y/2H2O8.1. 26. 8.8812 x 26/44 = 7.2 g (C)2 x 8.88/18= 0.9 g (H)7.2+0.9= 8.17.2/12 : 0.9:1= 0.6:0.9= 1:1.5 = 4:6C4H6.

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9.35g XgC2H4 + 3O2 > 2CO2 + 2H2O   3*32g 2*18gA: m(theor H2O) = X = 9.35*2*18/(3*32) = 3.506g.B: η(H2O) = m(theor H2O)/m(pract H2O) = 2.63g/3.506g*100% = 75.01% = 75%.

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At first the balanced chemical equation should be written:2C6H6+15O2=12CO2+6H2O.It is seen that heat here should be written for 2 moles of benzene. So the given number of moles should be...

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If 10mg of benzene is subjects to combustion analysis, what mass of CO2 and H2O will be produced?

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