A compound contains N=25.94% and O=74.06%. If its vapour density is 54.2, calculate its molecular formula.
(A) $NO$
(B) $N{{O}_{2}}$
(C) ${{N}_{2}}O$
(D) ${{N}_{2}}{{O}_{5}}$
Answer
624k+ views
Hint: Molecular mass is $2\times $Vapour density. Molecular formula is whole number multiple of empirical formula. Empirical formula is the simplest whole number ratio of various atoms present in a compound. mass % of an element is the ratio of mass of that element in compound to molar mass of compound.
Complete solution step by step:
-Empirical formula is the simplest whole number ratio of various atoms present in a compound.
-Molecular formula shows exact number of different types of atoms present in a molecule of a compound.
-Empirical formula can be calculated If percent mass of element is known. If molar mass is known, molecular formula can be calculated.
- A compound contains N=25.94% and O=74.06%. so 100g of compound will contain 25.94g of N and 74.06 g of O.
-As atomic mass of nitrogen is 14 g per mole and atomic mass of oxygen is 16 g per mole.
Number of moles of N= $=\dfrac{25.94}{14}=1.85$
Number of moles of O= $\dfrac{74.06}{16}=4.63$
The mole ratio of nitrogen to oxygen is 2: 5 as numbers can be rounded off to make whole numbers.
Hence, the empirical formula is ${{N}_{2}}{{O}_{5}}$. Empirical mass =28+80=108g
Molecular mass= $2\times $vapour density=108.4
Molecular formula= $n\times $ Empirical formula
$n=\dfrac{molecular\text{ mass}}{empirical\text{ mass}}=\dfrac{108.4}{108}=1$
Hence, Molecular formula=Empirical formula
A compound contains N=25.94% and O=74.06%. If its vapour density is 54.2, its molecular formula is
(D) ${{N}_{2}}{{O}_{5}}$
Note: molecular formula can be determined also by this method. As we know molecular mass is twice of vapour density. Hence Molecular mass is approximately 108g, only option (D) has molecular mass as 108g. Empirical formula is the simplest whole number ratio of various atoms present in a compound. mass % of an element is the ratio of mass of that element in compound to molar mass of compound.
Complete solution step by step:
-Empirical formula is the simplest whole number ratio of various atoms present in a compound.
-Molecular formula shows exact number of different types of atoms present in a molecule of a compound.
-Empirical formula can be calculated If percent mass of element is known. If molar mass is known, molecular formula can be calculated.
- A compound contains N=25.94% and O=74.06%. so 100g of compound will contain 25.94g of N and 74.06 g of O.
-As atomic mass of nitrogen is 14 g per mole and atomic mass of oxygen is 16 g per mole.
Number of moles of N= $=\dfrac{25.94}{14}=1.85$
Number of moles of O= $\dfrac{74.06}{16}=4.63$
The mole ratio of nitrogen to oxygen is 2: 5 as numbers can be rounded off to make whole numbers.
Hence, the empirical formula is ${{N}_{2}}{{O}_{5}}$. Empirical mass =28+80=108g
Molecular mass= $2\times $vapour density=108.4
Molecular formula= $n\times $ Empirical formula
$n=\dfrac{molecular\text{ mass}}{empirical\text{ mass}}=\dfrac{108.4}{108}=1$
Hence, Molecular formula=Empirical formula
A compound contains N=25.94% and O=74.06%. If its vapour density is 54.2, its molecular formula is
(D) ${{N}_{2}}{{O}_{5}}$
Note: molecular formula can be determined also by this method. As we know molecular mass is twice of vapour density. Hence Molecular mass is approximately 108g, only option (D) has molecular mass as 108g. Empirical formula is the simplest whole number ratio of various atoms present in a compound. mass % of an element is the ratio of mass of that element in compound to molar mass of compound.
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