(B) Mass of metal $(M) = 1.25 \ g$Mass of metal oxide $= 1.68 \ g$Mass of oxygen $= 1.68 \ g - 1.25 \ g = 0.43 \ g$Moles of $M = \frac{1.25 \ g}{69.7

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(B) Mass of metal $(M) = 1.25 \ g$Mass of metal oxide $= 1.68 \ g$Mass of oxygen $= 1.68 \ g - 1.25 \ g = 0.43 \ g$Moles of $M = \frac{1.25 \ g}{69.7 \ g \ mol^{-1}} \approx 0.0179 \ mol$Moles of $O = \frac{0.43 \ g}{16.0 \ g \ mol^{-1}} \approx 0.0269 \ mol$Ratio of moles $(M:O) = 0.0179 : 0.0269$Dividing by the smallest value $(0.0179)$: $1 : 1.502 \approx 1 : 1.5$Converting to whole numbers by multiplying by $2$: $2 : 3$$\therefore$ The empirical formula is $M_2O_3$.

Class 11 Chemistry Some Basic Concepts of Chemistry Percentage composition and Molecular formula

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$1.25 \ g$ of a metal $(M)$ reacts with oxygen completely to produce $1.68 \ g$ of metal oxide. The empirical formula of the metal oxide is
[molar mass of $M$ and $O$ are $69.7 \ g \ mol^{-1}$ and $16.0 \ g \ mol^{-1}$,respectively]

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A
$M_2O$
B
$M_2O_3$
C
$MO_2$
D
$M_3O_4$

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Percentage composition and Molecular formula

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$1.25 \ g$ of a metal $(M)$ reacts with oxygen completely to produce $1.68 \ g$ of metal oxide. The empirical formula of the metal oxide is[molar mass of $M$ and $O$ are $69.7 \ g \ mol^{-1}$ and $16.0 \ g \ mol^{-1}$,respectively]

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The empirical formula of a compound is $CH$. Its molecular weight is $78$. The molecular formula of the compound will be

$A$ compound has the following composition by weight: $Na = 18.60 \%$,$S = 25.80 \%$,$H = 4.02 \%$,and $O = 51.58 \%$. Assuming that all the hydrogen atoms in the compound are part of water of crystallisation,the correct molecular formula of the compound is:

$A$ compound contains $0.25\%$ of a metal with an atomic weight of $59$. If the compound contains one atom of the metal per molecule,what is the minimum molecular weight of the compound?

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$1.25 \ g$ of a metal $(M)$ reacts with oxygen completely to produce $1.68 \ g$ of metal oxide. The empirical formula of the metal oxide is
[molar mass of $M$ and $O$ are $69.7 \ g \ mol^{-1}$ and $16.0 \ g \ mol^{-1}$,respectively]