Question

Complete combustion of 0.0858 g 0f compound X gives 2.64 g of ${CO}_2$ and 1.26 g of $H_{2}O$. The lowest molecular mass X can have:

Answer 1

Given

molar mass of $C{o}_2=2.630g$ or since the molar mass of $C{o}_2$ is 44g/mol

we have $0.05976$ moles of $C{o}_2$ since there is 1 mole. so C in $C{o}_2$

we have $0.05976$ moles of C or $0.718$ g of C

mass of $H_{2}O$=$1.280$ g or since the molar mass of water is 18 g/mol

we have $0.07105$ mole of $H_{2}o$ since there are 2 moles of H in 1 moles of $H_{2}o$=$0.14210$ moles of H or $0.143$ g of H molar ratio $C:H$=$0.05976:0.14210$

mass of $C+H$=$0.861$ g

or, after divinding by the smallest $0.05976$

molar ratio of $C:H$=$1.000:2.378$

multiply by 3 to get the whole number $3.000$ $7.134$

empirical formula is $C_3{H}_7$

minimum molar mass is $43g/mol$