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Question Number 203878 by Spillover last updated on 31/Jan/24

$$\mathrm{A}\mathrm{compound}\mathrm{M}\mathrm{is}\mathrm{composed}\mathrm{of}52.2\mathrm{\%}$$$$\mathrm{carbon},13\mathrm{\%}\mathrm{hydrogenand}\mathrm{the}\mathrm{rest}$$$$\mathrm{is}\mathrm{oxygen}.\mathrm{if}\mathrm{the}\mathrm{molecuar}\mathrm{mass}\mathrm{of}\mathrm{M}\mathrm{is}138$$$$\left(\mathrm{a}\right)\mathrm{The}\mathrm{empirical}\mathrm{formular}$$$$\left(\mathrm{b}\right)\mathrm{The}\mathrm{molecular}\mathrm{formular}$$$$31/1/2024$$

Answered by Calculusboy last updated on 31/Jan/24

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:\mathit{t}\mathit{o}\mathit{g}\mathit{e}\mathit{t}\mathit{o}\mathit{x}\mathit{y}\mathit{g}\mathit{e}\mathit{n};100-(52.2+13)=34.8$$$$\mathit{e}\mathit{m}\mathit{p}\mathit{i}\mathit{r}\mathit{i}\mathit{c}\mathit{a}\mathit{l}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}:\underset{\frac{52.2}{12}}{\mathit{c}}\underset{\frac{13}{1}}{\mathit{h}}\underset{\frac{34.8}{16}}{\mathit{o}}$$$$4.35132.175$$$$\mathit{d}\mathit{i}\mathit{v}\mathit{i}\mathit{d}\mathit{e}\mathit{b}\mathit{y}\mathit{s}\mathit{m}\mathit{a}\mathit{l}\mathit{l}\mathit{e}\mathit{r}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}$$$$25.9771$$$$\mathit{e}\mathit{m}\mathit{p}\mathit{i}\mathit{r}\mathit{i}\mathit{c}\mathit{a}\mathit{l}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}:{\mathit{C}}_2{\mathit{H}}_6\mathit{O}$$$$\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{c}\mathit{u}\mathit{l}\mathit{a}\mathit{r}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}:{\left[{\mathit{C}}_2{\mathit{H}}_6\mathit{O}\right]}_{\mathit{n}}=138$$$${[\left(12\times 2\right)+\left(1\times 6\right)+\left(16\times 1\right)]}_{\mathit{n}}=138$$$${[24+6+16]}_{\mathit{n}}=138$$$$46\mathit{n}=138$$$$\mathit{n}=\frac{138}{46}=3$$$$\therefore \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{c}\mathit{u}\mathit{l}\mathit{a}\mathit{r}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}:{\left[{\mathit{C}}_2{\mathit{H}}_6\mathit{O}\right]}_3={\mathit{C}}_6{\mathit{H}}_{18}{\mathit{O}}_3$$$$$$$$$$

Commented by Spillover last updated on 11/Feb/24

$$correct$$

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