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Question
question 23
1 pts
for a compound whose empirical formula has been determined to be ( a_xb_y ), the
molecular formula is obtained by multiplying each subscript by the number of
formula units per molecule or ( n ):
( (a_xb_y)_n = a_{nx}b_{ny} )
suppose a compound has been determined to have the empirical formula ( ch )
and a has a molecular mass of ( 78.12 mathrm{~g} / mathrm{mol} ). calculate ( n ), the number of formula
units using the following formula:
( \frac{\text { molecular or molar mass }left(mathrm{amu} \text { or } \frac{mathrm{g}}{mathrm{mol}}
ight)}{\text { empirical formula mass }left(mathrm{amu} \text { or } \frac{mathrm{g}}{mathrm{mol}}
ight)}=n \text { formula units/molecule } )
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Step1: Calculate the empirical formula mass
The empirical formula is \(CH\). The atomic mass of \(C = 12.01\space g/mol\) and \(H=1.01\space g/mol\).
The empirical formula mass \(M_{empirical}=(12.01 + 1.01)\space g/mol=13.02\space g/mol\)
Step2: Calculate \(n\)
We know that \(n=\frac{\text{molecular mass}}{\text{empirical formula mass}}\)
Given molecular mass \(M_{molecular} = 78.12\space g/mol\)
\(n=\frac{78.12\space g/mol}{13.02\space g/mol}\)
\(n = 6\)
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