Moles and masses

[Гост]

In the combustion of an unknown organic compound containing only carbon and hydrogen, there are 6.601 g CO2 and 2.252 g H2O. Determine the empirical formula of unknown.

Why the answer is not 5+5=10 g? Conservation of mass is not working here??
And if someone can explain me how surplus and shortage are working, i will be super grateful!

[Гост]

Oh, sorry i was talking about this problem: How many g of H2O will you get from 5g of H2 and 5g of O2

[Гост]

Oh, sorry i was talking about this problem: How many g of H2O will you get from 5g of H2 and 5g of O2

[ammornil]

Oh, sorry i was talking about this problem: How many g of H2O will you get from 5g of H2 and 5g of O2

Assuming that the provided masses are for quantities of molecules of Hydrogen and Oxygen, and we accept the approximations [tex]M_{H}=1[g/mol], \quad M_{O}=16[g/mol][/tex]:
[tex]\\[12pt] m_{H_{2}}=5[g], \quad M_{H_{2}}=2\cdot{}M_{H}=2[g/mol], \quad \\m_{O_{2}}=5[g], \quad M_{O_{2}}=2\cdot{}M_{O}=32[g/mol] \\ M_{H_{2}O} = 2\cdot{}M_{H} + M_{O} = 18[g/mol] \\[12pt] 2H_{2} + O_{2} \longrightarrow 2H_{2}O \quad \Leftrightarrow \quad n(H_{2}):n(O_{2}):n(H_{2}O)=2:1:2 \quad \Rightarrow \quad \begin{cases} n(H_{2}):n(H_{2}O)=1:1 \\ n(O_{2}):n(H_{2}O)=1:2 \end{cases} \\[12pt] n_{H_{2}}=\dfrac{m_{H_{2}}}{M_{H_{2}}}=\dfrac{5}{2}[mol] \\ n_{O_{2}}=\dfrac{m_{O_{2}}}{O_{H_{2}}}=\dfrac{5}{32}[mol] \\[12pt] n_{H} > n_{O} \Rightarrow n_{H_{2}O}=2\cdot{n_{O_{2}}}=\frac{5}{16}[mol] \quad \Rightarrow \quad m_{H_{2}O}=M_{H_{2}O}\cdot{}n_{H_{2}O}=18\cdot{}\frac{5}{16}=\dfrac{45}{8}=5\frac{5}{8}[g][/tex]

NOTE:

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There will be leftover Hydrogen.
[tex]\frac{5}{16}[mol](H_{2}) + \frac{5}{32}[mol](O_{2}) = \frac{5}{16}[mol](H_{2}O)\\[6pt] n_{\text{leftover} H_{2}}=\frac{5}{2}-\frac{5}{16}=\frac{35}{16}[mol][/tex]