
In an organic compound of molar mass greater than \[100\] containing only $C,H$ and $N$ , the percentage of $C$ is $6$ times the percentage of $H$ while the sum of the percentages of $C$ and $H$ is $1.5$ times the percentage of $N$. What is the least molar mass?
A. $175$
B. $140$
C. $105$
D. $210$
Answer
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Hint: Molecules and atoms are very tiny substances both in size and mass. Every element in the chemistry consists of atomic number and atomic mass. Every element shows different atomic mass and number and therefore represented accordingly in the periodic table.
Complete step-by-step answer:
Let us assume that compound is ${{C}_{x}}{{H}_{y}}{{N}_{z}}$
We know that atomic number of $C=12,H=1,N=14$
Now we will substitute these values to calculate the mass,
Hence the mass of carbon $=12x$
Mass of hydrogen $=y$
Mass of nitrogen $=14z$
Therefore total mass $=12x+y+14z$
Now, we will calculate the percentage of carbon $=\dfrac{12x}{12x+y+14z}\times 100$
It is given that,
Mass of percentage of carbon $=6\times $ mass of percentage of hydrogen
Now we will substitute the above values we get,
$\Rightarrow 12x=6y$
$\Rightarrow y=2x$
Mass of percentage of carbon $+$ mass of percentage of hydrogen $=1.5$ (mass of percentage of nitrogen)
Now we will substitute the above value we get,
$\Rightarrow 12x+y=1.5\times 14z$
$\Rightarrow 12x+2x=21z$
On further solving we get,
$\Rightarrow x=1.5z$
And $y=3z$
So therefore, the compound can be expressed as
${{C}_{3z}}{{H}_{6z}}{{N}_{2z}}$
Now choosing $z=1$ we get ${{C}_{3}}{{H}_{6}}{{N}_{2}}$
The molecular weight will be $=3\times 12+6\times 1+2\times 14$
The molecular weight $=70gmo{{l}^{-1}}$
But here it is given that the molecular weight is more $100$
Now choosing $z=1$we get ${{C}_{6}}{{H}_{12}}{{N}_{4}}$
The molecular weight will be $=6\times 12+12\times 1+4\times 14$
The molecular weight $=140gmo{{l}^{-1}}$
Here, the molecular weight is more than $100$
The least molar mass is $140gmo{{l}^{-1}}$
Note: Molar mass of the compound is defined as the addition of atomic mass of all the elements present in the compound. It is calculated in $gmo{{l}^{-1}}$ . relative molar mass is defined as the smallest mass unit of a compound with one twelfth of the mass of $C-12$ atom.
Complete step-by-step answer:
Let us assume that compound is ${{C}_{x}}{{H}_{y}}{{N}_{z}}$
We know that atomic number of $C=12,H=1,N=14$
Now we will substitute these values to calculate the mass,
Hence the mass of carbon $=12x$
Mass of hydrogen $=y$
Mass of nitrogen $=14z$
Therefore total mass $=12x+y+14z$
Now, we will calculate the percentage of carbon $=\dfrac{12x}{12x+y+14z}\times 100$
It is given that,
Mass of percentage of carbon $=6\times $ mass of percentage of hydrogen
Now we will substitute the above values we get,
$\Rightarrow 12x=6y$
$\Rightarrow y=2x$
Mass of percentage of carbon $+$ mass of percentage of hydrogen $=1.5$ (mass of percentage of nitrogen)
Now we will substitute the above value we get,
$\Rightarrow 12x+y=1.5\times 14z$
$\Rightarrow 12x+2x=21z$
On further solving we get,
$\Rightarrow x=1.5z$
And $y=3z$
So therefore, the compound can be expressed as
${{C}_{3z}}{{H}_{6z}}{{N}_{2z}}$
Now choosing $z=1$ we get ${{C}_{3}}{{H}_{6}}{{N}_{2}}$
The molecular weight will be $=3\times 12+6\times 1+2\times 14$
The molecular weight $=70gmo{{l}^{-1}}$
But here it is given that the molecular weight is more $100$
Now choosing $z=1$we get ${{C}_{6}}{{H}_{12}}{{N}_{4}}$
The molecular weight will be $=6\times 12+12\times 1+4\times 14$
The molecular weight $=140gmo{{l}^{-1}}$
Here, the molecular weight is more than $100$
The least molar mass is $140gmo{{l}^{-1}}$
Note: Molar mass of the compound is defined as the addition of atomic mass of all the elements present in the compound. It is calculated in $gmo{{l}^{-1}}$ . relative molar mass is defined as the smallest mass unit of a compound with one twelfth of the mass of $C-12$ atom.
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