Determine the empirical fromula of kelvar, used in making bullet proof vests, is `70.6% C,4.2% H,11.8%N and 13.4%O`:

Options

A. `C_(7)H_(5)NO_(2)`
B. `C_(7)H_(5)N_(2)O`
C. `C_(7)H_(9)NO`
D. `C_(7)H_(5)NO`

Answer

To determine the empirical formula of Kevlar, we will follow these steps:

### Step 1: Convert percentages to grams
Assume we have 100 grams of the compound. This means we can directly use the percentages as grams:
- Carbon (C): 70.6 g
- Hydrogen (H): 4.2 g
- Nitrogen (N): 11.8 g
- Oxygen (O): 13.4 g

### Step 2: Convert grams to moles
Next, we convert the mass of each element to moles using their atomic masses:
- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Hydrogen (H) = 1 g/mol
- Molar mass of Nitrogen (N) = 14 g/mol
- Molar mass of Oxygen (O) = 16 g/mol

Calculating the moles for each element:
- Moles of Carbon (C) = 70.6 g / 12 g/mol = 5.8833 moles
- Moles of Hydrogen (H) = 4.2 g / 1 g/mol = 4.2 moles
- Moles of Nitrogen (N) = 11.8 g / 14 g/mol = 0.8429 moles
- Moles of Oxygen (O) = 13.4 g / 16 g/mol = 0.8375 moles

### Step 3: Find the smallest number of moles
Identify the smallest number of moles among the calculated values:
- The smallest value is for Nitrogen (N) = 0.8429 moles.

### Step 4: Divide by the smallest number of moles
Now, divide each mole value by the smallest number of moles to find the ratio:
- C: 5.8833 / 0.8429 ≈ 6.98 ≈ 7
- H: 4.2 / 0.8429 ≈ 4.97 ≈ 5
- N: 0.8429 / 0.8429 = 1
- O: 0.8375 / 0.8429 ≈ 0.99 ≈ 1

### Step 5: Write the empirical formula
Now we can write the empirical formula using the ratios obtained:
- The empirical formula is C7H5N1O1 or simply C7H5NO.

### Final Answer:
The empirical formula of Kevlar is **C7H5NO**.

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