Question

lets say an unknown substance in the lab is shown to be composed of 36.84% nitrogen and 63.16% oxygen by mass. it has a molecular weight of 152 amu. can you use this information to determine what the molecular formula of this substance would be?
to find the molecular formula, well need to compare the moles of each of the elements in the substance. right now, you only have mass values (in grams) for each of the elements.
assuming you had 100 grams of the substance, convert the percent mass values of the elements into grams.
n = 36.8% g
o = 63.2% g

Explanation:

Step1: Calculate the mass of each element

Assume 100g of the substance.
For nitrogen ($N$):
$N = 36.84g$ (since 36.84% of 100g is $0.3684\times100 = 36.84g$)
For oxygen ($O$):
$O = 63.16g$ (since 63.16% of 100g is $0.6316\times100 = 63.16g$)

Step2: Calculate the moles of each element

The molar mass of $N$ is $M_N=14.01g/mol$, and the molar mass of $O$ is $M_O = 16.00g/mol$.
Moles of $N$: $n_N=\frac{m_N}{M_N}=\frac{36.84g}{14.01g/mol}\approx2.63mol$
Moles of $O$: $n_O=\frac{m_O}{M_O}=\frac{63.16g}{16.00g/mol}\approx3.95mol$

Step3: Find the ratio of moles

Divide each number of moles by the smaller number of moles ($n_N = 2.63mol$)
Ratio of $N$: $\frac{2.63}{2.63}=1$
Ratio of $O$: $\frac{3.95}{2.63}\approx1.5$
Multiply by 2 to get whole - number ratio. So the empirical formula is $N_2O_3$ (empirical formula weight $M_{EF}=2\times14.01 + 3\times16.00=28.02+48.00 = 76.02g/mol$)

Step4: Calculate the multiple ($n$)

The molecular weight $M_{MW}=152g/mol$
$n=\frac{M_{MW}}{M_{EF}}=\frac{152g/mol}{76.02g/mol}\approx2$

Step5: Determine the molecular formula

Multiply the empirical formula by $n$
Molecular formula: $(N_2O_3)\times2=N_4O_6$

Answer:

The molecular formula of the substance is $N_4O_6$