
A compound contains 38.8% C, 16.0% H and 45.2% N. The empirical formula of the compound would be:
A.$C{H_3}N{H_2} $
B.$C{H_3}CN $
C.${C_2}{H_5}CN $
D.$C{H_2}{\left( {NH} \right)_2}$
Answer
588k+ views
Hint: The empirical formula for a compound is defined as the ratio of the numbers of atoms of different elements that are combined in order to form the molecule of the compound. For example, if we have element X, Y and Z then their formula will be their ratio X:Y:Z.
Complete step by step answer:
Mass of Nitrogen = \[45.2{\text{ }}g\]
Mass of Carbon = \[38.8{\text{ }}g\]
Mass of Hydrogen = \[16{\text{ }}g\]
In order to find the empirical formula of the compound, the moles of each element has to be calculated;
The formula used for empirical formula calculation is;
\[Moles{\text{ }}of{\text{ }}atom{\text{ }} = \dfrac{{mass{\text{ of }}atom}}{{molar{\text{ mass of }}atom{\text{ }}}}............(1)\]
Substituting the given values in equation 1;
For carbon;
Moles of carbon \[{\text{ }} = \dfrac{{mass{\text{ }}of{\text{ }}carbon{\text{ }}}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}carbon{\text{ }}}}\]
Moles of carbon \[{\text{ }} = \dfrac{{{\text{38}}{\text{.8g }}}}{{12g/mole{\text{ }}}}\]
Moles of carbon \[{\text{ }} = 3.23moles\]
For hydrogen;
Moles of hydrogen \[{\text{ }} = \dfrac{{{\text{16g }}}}{{1g/mole{\text{ }}}}\]
Moles of hydrogen \[{\text{ }} = 1moles\]
For nitrogen;
Moles of Nitrogen \[{\text{ }} = \dfrac{{{\text{45}}{\text{.2g }}}}{{14g/mole{\text{ }}}}\]
Moles of Nitrogen \[ = 3.23moles\]
Now, the next step will be to divide the moles calculated for each element with the smallest mole value i.e. 3.32.
Thus for carbon element;
Ratio will be = $\dfrac{{3.23}}{{3.23}} = 1$
For nitrogen, the ratio will be= $\dfrac{{3.23}}{{3.23}} = 1$
For hydrogen, the ratio will be = $\dfrac{{16}}{{3.23}} = 5$
According to the calculations, the ratio is for carbon, hydrogen and nitrogen is= 1:5:1
Thus the formula of the compound is$C{H_5}N$.
So, the correct answer is “Option A”.
Note:
Mass of an atom is due to the weight of neutrons and protons in its nucleus, the weight is the same for both but one is charged and the other is neutral.
Mole is defined as the exactly equal to \[6.02214076 \times {10^{23}}\;\]particles in an element, and is the unit of measurement for the amount of a substance present.
Complete step by step answer:
Mass of Nitrogen = \[45.2{\text{ }}g\]
Mass of Carbon = \[38.8{\text{ }}g\]
Mass of Hydrogen = \[16{\text{ }}g\]
In order to find the empirical formula of the compound, the moles of each element has to be calculated;
The formula used for empirical formula calculation is;
\[Moles{\text{ }}of{\text{ }}atom{\text{ }} = \dfrac{{mass{\text{ of }}atom}}{{molar{\text{ mass of }}atom{\text{ }}}}............(1)\]
Substituting the given values in equation 1;
For carbon;
Moles of carbon \[{\text{ }} = \dfrac{{mass{\text{ }}of{\text{ }}carbon{\text{ }}}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}carbon{\text{ }}}}\]
Moles of carbon \[{\text{ }} = \dfrac{{{\text{38}}{\text{.8g }}}}{{12g/mole{\text{ }}}}\]
Moles of carbon \[{\text{ }} = 3.23moles\]
For hydrogen;
Moles of hydrogen \[{\text{ }} = \dfrac{{{\text{16g }}}}{{1g/mole{\text{ }}}}\]
Moles of hydrogen \[{\text{ }} = 1moles\]
For nitrogen;
Moles of Nitrogen \[{\text{ }} = \dfrac{{{\text{45}}{\text{.2g }}}}{{14g/mole{\text{ }}}}\]
Moles of Nitrogen \[ = 3.23moles\]
Now, the next step will be to divide the moles calculated for each element with the smallest mole value i.e. 3.32.
Thus for carbon element;
Ratio will be = $\dfrac{{3.23}}{{3.23}} = 1$
For nitrogen, the ratio will be= $\dfrac{{3.23}}{{3.23}} = 1$
For hydrogen, the ratio will be = $\dfrac{{16}}{{3.23}} = 5$
According to the calculations, the ratio is for carbon, hydrogen and nitrogen is= 1:5:1
Thus the formula of the compound is$C{H_5}N$.
So, the correct answer is “Option A”.
Note:
Mass of an atom is due to the weight of neutrons and protons in its nucleus, the weight is the same for both but one is charged and the other is neutral.
Mole is defined as the exactly equal to \[6.02214076 \times {10^{23}}\;\]particles in an element, and is the unit of measurement for the amount of a substance present.
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