
The relative number of atoms of elements X and Y in a compound is 0.25 and 0.5. The empirical formula of compound is:
A ) \[{\text{XY}}\]
B ) \[{{\text{X}}_2}{\text{Y}}\]
C ) \[{\text{X}}{{\text{Y}}_2}\]
D ) \[{{\text{X}}_2}{{\text{Y}}_2}\]
Answer
571.8k+ views
Hint: Calculate the ratio of the relative number of atoms of element X to the relative number of atoms of element Y. For this divide 0.25 with 0.5. 0.25 is the relative number of atoms of element X and 0.5 is the relative number of atoms of element Y.
Complete step by step answer:
Let the empirical formula of the compound be \[{{\text{X}}_a}{{\text{Y}}_b}\].
In the empirical formula, a and b represents the relative number of atoms of X and Y respectively. Write the ratio of the relative number of atoms of element X to the relative number of atoms of element Y.
\[\dfrac{{{\text{The relative number of atoms of element X}}}}{{{\text{The relative number of atoms of element Y}}}} = \dfrac{a}{b}\]
The relative number of atoms of elements X and Y in a compound is 0.25 and 0.5.
The ratio of the relative number of atoms of element X to the relative number of atoms of element Y will be
\[\dfrac{{{\text{The relative number of atoms of element X}}}}{{{\text{The relative number of atoms of element Y}}}} = \dfrac{a}{b} \\
= \dfrac{{0.25}}{{0.5}} \\
= \dfrac{1}{2} \\\]
Rearrange the above ratio.
\[{\text{The relative number of atoms of element Y}} = 2 \times {\text{The relative number of atoms of element X}}\]
This is the relative number of atoms of element X to the relative number of atoms of element Y. You can conclude that the number of atoms of element Y is twice the number of atoms of element X.
Hence, the empirical formula of compound is \[{\text{X}}{{\text{Y}}_2}\].
Hence, option C ) \[{\text{X}}{{\text{Y}}_2}\] is the correct option.
Note: Empirical formula gives the ratio of atoms of different elements present in a compound. This ratio is the simplest positive integer ratio. Empirical formula can be obtained from the molecular formula by dividing with an integer. For example, the molecular formula of ethane is \[{{\text{C}}_2}{{\text{H}}_6}\] but its empirical formula is \[{\text{C}}{{\text{H}}_3}\].
Complete step by step answer:
Let the empirical formula of the compound be \[{{\text{X}}_a}{{\text{Y}}_b}\].
In the empirical formula, a and b represents the relative number of atoms of X and Y respectively. Write the ratio of the relative number of atoms of element X to the relative number of atoms of element Y.
\[\dfrac{{{\text{The relative number of atoms of element X}}}}{{{\text{The relative number of atoms of element Y}}}} = \dfrac{a}{b}\]
The relative number of atoms of elements X and Y in a compound is 0.25 and 0.5.
The ratio of the relative number of atoms of element X to the relative number of atoms of element Y will be
\[\dfrac{{{\text{The relative number of atoms of element X}}}}{{{\text{The relative number of atoms of element Y}}}} = \dfrac{a}{b} \\
= \dfrac{{0.25}}{{0.5}} \\
= \dfrac{1}{2} \\\]
Rearrange the above ratio.
\[{\text{The relative number of atoms of element Y}} = 2 \times {\text{The relative number of atoms of element X}}\]
This is the relative number of atoms of element X to the relative number of atoms of element Y. You can conclude that the number of atoms of element Y is twice the number of atoms of element X.
Hence, the empirical formula of compound is \[{\text{X}}{{\text{Y}}_2}\].
Hence, option C ) \[{\text{X}}{{\text{Y}}_2}\] is the correct option.
Note: Empirical formula gives the ratio of atoms of different elements present in a compound. This ratio is the simplest positive integer ratio. Empirical formula can be obtained from the molecular formula by dividing with an integer. For example, the molecular formula of ethane is \[{{\text{C}}_2}{{\text{H}}_6}\] but its empirical formula is \[{\text{C}}{{\text{H}}_3}\].
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

