Answer
Simplest formula of timolol is $C_{13}H_{24}N_{4}O_{3}S$
Molecular formula of timolol is $C_{13}H_{24}N_{4}O_{3}S$
Work Step by Step
First, we need to calculate the mass of each element in 3.16g timolol using the percent composition:
Mass of C=$total mass\times percent mass=3.16\times\frac{49.4}{100}=1.56104g$
Mass of H=$total mass\times percent mass=3.16\times\frac{7.64}{100}=0.241424g$
Mass of N=$total mass\times percent mass=3.16\times\frac{17.7}{100}=0.55932g$
Mass of O=$total mass\times percent mass=3.16\times\frac{15.2}{100}=0.48032g$
Mass of S=$total mass\times percent mass=3.16\times\frac{10.1}{100}=0.31916g$
Then we calculate the number of moles of each element using the formula:
Number of moles=$\frac{mass}{moleculat mass}$
Number of moles of C=$\frac{1.56104}{12}\approx0.13 amu$
Number of moles of H=$\frac{0.241424}{1}\approx0.24 amu$
Number of moles of N=$\frac{0.55932}{14}\approx0.04 amu$
Number of moles of O=$\frac{0.48032}{12}\approx0.03 amu$
Number of moles of S=$\frac{0.31916}{32}\approx0.01 amu$
Finally, we calculate the ratio of elements to the lowest number of mole element.
The ratio of C to S$=\frac{0.13}{0.01}=13$
The ratio of H to S$=\frac{0.24}{0.01}=24$
The ratio of N to S$=\frac{0.04}{0.01}=4$
The ratio of O to S$=\frac{0.03}{0.01}=3$
The simplest formula is $C_{13}H_{24}N_{4}O_{3}S$
The molar mass of timolol can be calculated using the formula:
Molar mass$=\frac{mass}{number of moles}=\frac{3.16}{0.001}=316 amu$
The simplest formula of timolol molar mass$=13\times12+24\times1+4\times14+3\times16+1\times32=316amu$
Since the simplest formula of timolol molar mass is similar to timolol molar mass So, the simplest formula of timolol is the molecular formula of timolol $C_{13}H_{24}N_{4}O_{3}S$.
