Question

Sulfadiazine, a sulfur drug used in the treatment of bacterual infections, is found on analysis: 48.0% carbon, 4.0% hydrogen, 22.4% nitrogen, 12.8% sulfur, and 12.8% oxygen. The molecular mass was found to be 250 amu. Calculate the molecular formula for this drug.Let's complete step one, and assume we have 100 grams total of the substance -record the number of grams of each element we would have below:Grams carbon:__________Grams hydrogen:___________Grams nitrogen:___________Grams sulfur:___________Grams oxygen:__________Now, let's convert the number of grams of each elemen into moles below:_________________________Now, let's divide the number of moles of each element by the smallest number obtained:___________________________What is the empirical formula for this compound?____________Using the actual molecular mass of 250 amu, determine the molecular formula for this compound:_____________

Answer 1

In this case we are given the percentage of each element in a sample, and we know that the molecular mass is 250 amu.

The amu unit is defined as:

1/12 of the mass of an unbound neutral atom of carbon-12.

And 1 mol of uma equals 1g.

So in this case knowing that 1 molecules has a weight of 250 uma, we can calculate the molar mass (g/mol), which is 250g/mol.

Now, we start in step 1.

Assuming that we have 100g of the substance, using the percentage given we calculate the mass (mx) of each element using the following formula:

`m_x=(percentage_x).100g`
`\begin{gathered} m_C=48g \\ m_H=4g \\ m_N=22.4g \\ m_S=12.8g \\ m_O=12.8g \end{gathered}`

Step 2

To convert the number of grams of each element into moles we use the molar mass (M) of each of them, this information we obtain from the periodic table:

`\begin{gathered} M_C=12.01g/mol \\ M_H=1.0g/mol \\ M_N=14.0g/mol \\ M_S=32.06g/mol \\ M_O=16.g/mol \end{gathered}`

So we calculate the moles (n) as follows:

`n=(m)/(M)`

`\begin{gathered} n_C=4mol \\ n_H=4mol \\ n_N=1.6mol \\ n_S=0.35mol \\ n_O=0.8mol \end{gathered}`

Step 3

We divide the number of moles of each element by the smallest number obtained:

`\begin{gathered} C=(4mol)/(0.35mol)=11.43 \\ H=(4mol)/(0.35mol)=11.43 \\ N=(1.6mol)/(0.35mol)=4.57 \\ S=(0.35mol)/(0.35mol)=1 \\ O=(0.8mol)/(0.35mol)=2.28 \end{gathered}`

Steps 4 and 5

To calculate the empirical formula, we use