We read your photo as:

A 3.44 g sample of an oxide of nitrogen contains 30.45% nitrogen by mass. What is its empirical formula? If the molar mass of the oxide is 90 ± 5 g/mol, what is its molecular formula?

Chemistry

ScanSolve AI Answer

Answers

2 parts
a

NO2\text{NO}_2

b

N2O4\text{N}_2\text{O}_4

Step-by-step solution

6 steps

Here's how to solve .44:

Part a· 3 steps

  1. Assume a 100 g sample of the oxide.

    The mass of nitrogen (N) is 30.45 g.
    The mass of oxygen (O) is 100 g−30.45 g=69.55 g100 \text{ g} - 30.45 \text{ g} = 69.55 \text{ g}.

  2. Convert the mass of each element to moles using their molar masses (N = 14.01 g/mol, O = 16.00 g/mol).

    Moles of N=30.45 g14.01 g/mol≈2.173 mol\text{Moles of N} = \frac{30.45 \text{ g}}{14.01 \text{ g/mol}} \approx 2.173 \text{ mol} Moles of O=69.55 g16.00 g/mol≈4.347 mol\text{Moles of O} = \frac{69.55 \text{ g}}{16.00 \text{ g/mol}} \approx 4.347 \text{ mol}

  3. Divide each mole value by the smallest number of moles to find the simplest whole-number ratio.

    Ratio for N=2.173 mol2.173 mol=1\text{Ratio for N} = \frac{2.173 \text{ mol}}{2.173 \text{ mol}} = 1 Ratio for O=4.347 mol2.173 mol≈2.00\text{Ratio for O} = \frac{4.347 \text{ mol}}{2.173 \text{ mol}} \approx 2.00

    The empirical formula is NO2\boxed{\text{NO}_2}.

Part b

N2O4\text{N}_2\text{O}_4

  1. Calculate the empirical formula mass (EFM) for NO2\text{NO}_2.

    EFM=(1×14.01 g/mol)+(2×16.00 g/mol)=14.01+32.00=46.01 g/mol\text{EFM} = (1 \times 14.01 \text{ g/mol}) + (2 \times 16.00 \text{ g/mol}) = 14.01 + 32.00 = 46.01 \text{ g/mol}

  2. Determine the integer nn by dividing the given molar mass by the empirical formula mass.

    We'll use 90 g/mol for the molar mass, noting the ±5\pm 5 g/mol range.

    n=Molar MassEFM=90 g/mol46.01 g/mol≈1.956n = \frac{\text{Molar Mass}}{\text{EFM}} = \frac{90 \text{ g/mol}}{46.01 \text{ g/mol}} \approx 1.956

    This value is very close to 2. If we use a molar mass of 92 g/mol (which is within the 90±590 \pm 5 g/mol range), n=92 g/mol46.01 g/mol≈2.00n = \frac{92 \text{ g/mol}}{46.01 \text{ g/mol}} \approx 2.00. So, n=2n=2.

  3. Multiply the subscripts in the empirical formula by nn to get the molecular formula.

    Molecular Formula=(NO2)2=N2O4\text{Molecular Formula} = (\text{NO}_2)_2 = \boxed{\text{N}_2\text{O}_4}
Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Chemistry Questions

QThe mass of oxygen in a 20.0 g sample of sodium oxide is 8.2 g. Determine the empirical and…QUse the results to find the percentage mass of Magnesium & Oxygen in Magnesium oxide. Determine the…QGiven 30.45 mass % N, the remaining percentage is oxygen.QGiven molar mass (M) = 74 g⁻¹ and percentage of oxygen by mass = 43.24\%.QWhen 56.6 g of calcium and 30.5 g of nitrogen gas undergo a reaction that has a 93.0% yield, what…
See all Chemistry questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp