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Hint: Empirical Formula - A empirical formula is the formula that gives the simplest whole-number ratio of atoms in a compound. There are some definite steps for Determining an Empirical Formula. We have to start with the number of grams of each element, given in the problem.
A molecular formula consists of the mass of each element (chemical symbols) for the constituent elements present in the formula followed by numeric subscripts describing the number of atoms of each element present in the molecule.
Empirical measurements are based on a measurable (empirical) quantity like mass. benzene and acetylene have the same mass percent composition, and thus have the same ratio of elements to each other, that is, they have the same empirical formula molecular formula for benzene $,{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$ acetylene \[,{{\text{C}}_{\text{2}}}\text{H}{{ }_{2}}\]
Complete step by step answer:
Since the question already gave the empirical formula for both compounds, which is \[\text{CH}\]. Let's figure out the subscript, $n$ , in \[{{\left( \text{CH} \right)}_{\text{n}}}\] for their molecular formulas.
To do so, let's use the molar mass and divide by the formula mass of the empirical formula, \[\text{CH}\] , \[12+1=13\text{ g}.\]
Molecular formula $=n\times $ empirical formula
Molecular weight $=n\times $ empirical formula weight
For benzene:
subscript, \[n=\] molar mass benzene/formula mass of \[\text{CH}=\dfrac{78\text{g}}{13\text{ g}}=6\text{N}=6\] molecular formula for benzene \[,{{\left( \text{CH} \right)}_{6}},{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\]
For acetylene:
subscript, $n=$ molar mass acetylene/formula mass of \[\text{CH}=\dfrac{26\text{ g}}{13\text{ g}}=2\text{N}=2\] molecular formula for acetylene \[,{{\left( \text{CH} \right)}_{2}},{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\]
Note: Generally, The molecular formula shows the number of each type of atom in a molecule. In the structural formula, the structural formula shows the arrangement of the molecule. Different compounds can also have the same empirical formula. Samples are analyzed in specific elemental analysis tests to determine what percent of a particular element the sample is composed of.
A molecular formula consists of the mass of each element (chemical symbols) for the constituent elements present in the formula followed by numeric subscripts describing the number of atoms of each element present in the molecule.
Empirical measurements are based on a measurable (empirical) quantity like mass. benzene and acetylene have the same mass percent composition, and thus have the same ratio of elements to each other, that is, they have the same empirical formula molecular formula for benzene $,{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$ acetylene \[,{{\text{C}}_{\text{2}}}\text{H}{{ }_{2}}\]
Complete step by step answer:
Since the question already gave the empirical formula for both compounds, which is \[\text{CH}\]. Let's figure out the subscript, $n$ , in \[{{\left( \text{CH} \right)}_{\text{n}}}\] for their molecular formulas.
To do so, let's use the molar mass and divide by the formula mass of the empirical formula, \[\text{CH}\] , \[12+1=13\text{ g}.\]
Molecular formula $=n\times $ empirical formula
Molecular weight $=n\times $ empirical formula weight
For benzene:
subscript, \[n=\] molar mass benzene/formula mass of \[\text{CH}=\dfrac{78\text{g}}{13\text{ g}}=6\text{N}=6\] molecular formula for benzene \[,{{\left( \text{CH} \right)}_{6}},{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\]
For acetylene:
subscript, $n=$ molar mass acetylene/formula mass of \[\text{CH}=\dfrac{26\text{ g}}{13\text{ g}}=2\text{N}=2\] molecular formula for acetylene \[,{{\left( \text{CH} \right)}_{2}},{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\]
Note: Generally, The molecular formula shows the number of each type of atom in a molecule. In the structural formula, the structural formula shows the arrangement of the molecule. Different compounds can also have the same empirical formula. Samples are analyzed in specific elemental analysis tests to determine what percent of a particular element the sample is composed of.
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