A compound contains 4.07% hydrogen,24.27% carbon and 71.65% chlorine.Its molar mass is 98.96g. What are its empirical and molecular formulas?
Answers
the molecular formula
definition
A molecular formula tells you the ratios of each different element in a molecule. An empirical formula tells you just the lowest ratio of the different elements in each molecule. Each has a specific weight associated with it, and the molecular weight will always be a whole number multiple of the empirical weight. In this case, the empirical formula CH2O has an associated empirical weight of 12 + 2 + 16 = 30 g/equivalent. The molecular weight is (almost) exactly 4 times the empirical weight, meaning that the molecular formula is exactly four times the empirical formula: C4H8O4.
the empirical formula
definition
In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.[1] A simple example of this concept is that the empirical formula of sulphur monoxide, or SO, would simply be SO, as is the empirical formula of disulfur dioxide, S2O2. This means that sulfur monoxide and disulfur dioxide, both compounds of sulfur and oxygen, will have the same empirical formula. However, their chemical formulas, which express the number of atoms in each molecule of a chemical compound, may not be the same.
please mark me the brain list please i will be helpful for u and me please mark me please
Explanation:
Answer:
\LARGE{\bf {\underline{\underline{Given : }}}}
Given:
Percentage composition of Hydrogen in the compound is 4.07 %
Percentage composition of Carbon in the compound is 24.47 %
Percentage composition of Chlorine in the compound is 71.65 %
Molar mass = 99 g/mol
\LARGE{\underline{\underline{\bf{To \: find:}}}}
Tofind:
Molecular formula of the compound
{\LARGE{\underline{\underline{\bf{Solution:}}}}}
Solution:
Strep 1 : Calculating relative number of atoms
a. Hydrogen
Percentage composition = 4.07 %
Atomic weight of hydrogen = 1
Relative number of atoms = \sf{\dfrac{4.07}{1}}
1
4.07
= 4.07
b. Carbon
Percentage comppsition = 24.47 %
Aomic weight of carbon = 12
Relative number of atoms = \sf{\dfrac{24.47}{12}}
12
24.47
= 2.02
c. Chlorine
Percentage comppsition = 71.65%
Atomic weight of chlorine = 35.5
Relative number of atoms = \sf{\dfrac{71.65}{35.5}}
35.5
71.65
= 2.01
Step 2 : Calculating Simplest ratio for each element
Simplest ratio is calculted by dividing all the values we got by the smallest value we got.
The smallest value we got is 2.01
a. Hydrogen
Simplest ratio = \sf{\dfrac{4.07}{2.01}}
2.01
4.07
\approx≈ 2
b. Carbon
Simplest ratio = \sf{\dfrac{2.02}{2.01}}
2.01
2.02
\approx≈ 1
c. Chlorine
Simplest ratio = \sf{\dfrac{2.01}{2.01}}
2.01
2.01
= 1
The empirical formula becomes CH₂Cl
Molecular weight of Empirical formula = 12 + 2(1) + 35.5 = 49.5
We are already given that molecular weight as 99.
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Relation between Empirical formula and Molecular formula is given by ,
\star\large{\boxed{ \sf{ \purple{Molecular\:Formula=Empirical\:formula\times\:n}}}}⋆
MolecularFormula=Empiricalformula×n
n is some integer which is given by ,
\star\large{\boxed{\purple{ \sf{ n = \dfrac{Molecular\:Formula \: weight}{Empirical\:formula \: weight}}}}}⋆
n=
Empiricalformulaweight
MolecularFormulaweight
We know the values for calculating the value of n . So , by substituting we get ;
\begin{gathered} \\ : \implies \sf \: n = \frac{99}{49.5} \\ \\ \\ : \implies \sf {\boxed {\underline{ \sf{n = 2}}}}\end{gathered}
:⟹n=
49.5
99
:⟹
n=2
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Applying the first formulae , (i.e , relation between empirical formulae and Molecular formulae) we get ;
\begin{gathered} \\ \implies \sf Molecular\:Formula=Empirical\:formula \times \: n \\ \\ \\ \implies \sf Molecular\:Formula=CH_2Cl \times 2 \\ \\ \\ \implies \sf \: Molecular\:Formula=C_2H_4Cl_2\end{gathered}
⟹MolecularFormula=Empiricalformula×n
⟹MolecularFormula=CH
2
Cl×2
⟹MolecularFormula=C
2
H
4
Cl
2
Hence , The molecular formula of the given compound is C₂H₄Cl₂