Benzene, C6H6, reacts with nitric acid, HNO3. Two products are formed, one of which is water. The second product has a molar mass of 213 grams. The second product is composed of 33.8% carbon, 1.42% hydrogen, 19.7% nitrogen, and 45.1% oxygen. Write a balanced chemical equation for this reaction
Determine the empirical formula.
Take 100 g sample which gives you
33.8 g C
1.42 g H
19.7 g N
45.1 g O
-----------
Convert to mols.
33.8/12 = approx 2.8 C
1.42/1 = 1.42 H
19.7/14 = approx 1.41 N
45.1/16 = approx 2.8 O
-------------
Find the ratio
I think that is C2HNO2. The empirical mass is (24 + 1 + 14 + 32 = 71
molar mass in the problem is 213 so this must be 213/71 = about 3 and the molecular formula is C6H3N3O6 which I would bet was C6H3(NO2)3
C6H6 + HNO3 ==> H2O + C6H3(NO2)3
I will leave that for you to balance.
C6H6(g)+3HNO3(aq)=C6H3(NO2)3(aq)+3H2O(l)
To write a balanced chemical equation, we need to determine the reactants and products in the reaction.
Given:
Reactants:
1. Benzene, C6H6
2. Nitric acid, HNO3
Products:
1. Water, H2O
2. Second product with a molar mass of 213 grams and composition:
- Carbon: 33.8%
- Hydrogen: 1.42%
- Nitrogen: 19.7%
- Oxygen: 45.1%
Now, let's balance the equation step-by-step:
1. Start with the benzene, C6H6, and nitric acid, HNO3.
C6H6 + HNO3
2. The reaction produces water, H2O, and a second product.
C6H6 + HNO3 -> H2O + Second product
3. Determine the balanced equation for the second product.
The second product has a molar mass of 213 grams and the following composition percentages:
- Carbon: 33.8%
- Hydrogen: 1.42%
- Nitrogen: 19.7%
- Oxygen: 45.1%
Let's assume the chemical formula for the second product as CxHyNzOw.
To determine the subscripts (x, y, z, w), we can convert the composition percentages to grams. Then, we can convert the grams to moles using the molar masses of each element.
33.8% carbon:
Assuming 213 grams of the second product, the mass of carbon is given by:
mass of carbon = 33.8/100 * 213 grams = 71.894 grams
The molar mass of carbon (C) is approximately 12 grams/mol.
Moles of carbon = mass of carbon / molar mass of carbon = 71.894 grams / 12 grams/mol ≈ 5.991 moles
1.42% hydrogen:
Assuming 213 grams of the second product, the mass of hydrogen is given by:
mass of hydrogen = 1.42/100 * 213 grams = 3.02546 grams
The molar mass of hydrogen (H) is approximately 1 gram/mol.
Moles of hydrogen = mass of hydrogen / molar mass of hydrogen = 3.02546 grams / 1 gram/mol ≈ 3.02546 moles
19.7% nitrogen:
Assuming 213 grams of the second product, the mass of nitrogen is given by:
mass of nitrogen = 19.7/100 * 213 grams = 42.051 grams
The molar mass of nitrogen (N) is approximately 14 grams/mol.
Moles of nitrogen = mass of nitrogen / molar mass of nitrogen = 42.051 grams / 14 grams/mol ≈ 3.003 moles
45.1% oxygen:
Assuming 213 grams of the second product, the mass of oxygen is given by:
mass of oxygen = 45.1/100 * 213 grams = 95.763 grams
The molar mass of oxygen (O) is approximately 16 grams/mol.
Moles of oxygen = mass of oxygen / molar mass of oxygen = 95.763 grams / 16 grams/mol ≈ 5.985 moles
Based on these calculations, the subscripts (x, y, z, w) can be approximately determined as follows:
CxHyNzOw = C6.991H3.02546N3.003O5.985
To simplify the equation, we can multiply all subscripts by a common factor of approximately 33:
33C5.991H3.02546N3.003O5.985 ≈ C197.403H99.85058N99.099O197.805
Now, the balanced chemical equation can be written as:
C6H6 + HNO3 -> H2O + C197.403H99.85058N99.099O197.805
To write a balanced chemical equation for the reaction between benzene (C6H6) and nitric acid (HNO3), we need to understand the reaction and the products formed.
Given that one of the products is water (H2O), we can start by writing the reaction as follows:
C6H6 + HNO3 → H2O + ??
The second product has a molar mass of 213 grams and is composed of 33.8% carbon, 1.42% hydrogen, 19.7% nitrogen, and 45.1% oxygen.
To find the molecular formula of the second product, we need to determine the number of moles of each element present in the given percentages.
Let's assume we have 100 grams of the second product.
So, in 100 grams of the second product:
- Carbon (C) = 33.8 grams
- Hydrogen (H) = 1.42 grams
- Nitrogen (N) = 19.7 grams
- Oxygen (O) = 45.1 grams
Next, we need to convert these masses of each element to moles. To do this, we divide the mass of each element by its molar mass.
- Molar mass of carbon (C) = 12.01 g/mol
- Molar mass of hydrogen (H) = 1.01 g/mol
- Molar mass of nitrogen (N) = 14.01 g/mol
- Molar mass of oxygen (O) = 16.00 g/mol
Now, we can calculate the number of moles:
- Moles of carbon (C) = 33.8 g / 12.01 g/mol ≈ 2.81 mol
- Moles of hydrogen (H) = 1.42 g / 1.01 g/mol ≈ 1.41 mol
- Moles of nitrogen (N) = 19.7 g / 14.01 g/mol ≈ 1.40 mol
- Moles of oxygen (O) = 45.1 g / 16.00 g/mol ≈ 2.82 mol
The empirical formula indicates the ratios of elements in the simplest whole number form found in a compound. To find the empirical formula, we divide each of the moles by the smallest number of moles.
In this case, the smallest number of moles is approximately 1.40 mol, so dividing by this number gives:
- Moles of carbon (C) = 2.81 mol / 1.40 mol ≈ 2.01
- Moles of hydrogen (H) = 1.41 mol / 1.40 mol ≈ 1.01
- Moles of nitrogen (N) = 1.40 mol / 1.40 mol = 1.00
- Moles of oxygen (O) = 2.82 mol / 1.40 mol ≈ 2.01
Therefore, the empirical formula of the second product is approximately C2H1N1O2.
To determine the molecular formula, we need to find the ratio between the sum of the atomic masses of the empirical formula and the molar mass of the second product.
The molar mass of the empirical formula (C2H1N1O2) can be calculated as:
- (2 * atomic mass of carbon) + (1 * atomic mass of hydrogen) + (1 * atomic mass of nitrogen) + (2 * atomic mass of oxygen)
Substituting the respective values:
- (2 * 12.01) + (1 * 1.01) + (1 * 14.01) + (2 * 16.00) ≈ 78.04 g/mol
To find the molecular formula, we divide the molar mass of the second product (213 g) by the molar mass of the empirical formula (78.04 g/mol).
- Molecular formula = 213 g / 78.04 g/mol ≈ 2.73
Since the molecular formula must be a whole number, we round the value calculated to the nearest whole number:
- Molecular formula = 3
Hence, the molecular formula of the second product is C6H3N3O6.
Now, we can write the balanced chemical equation:
C6H6 + HNO3 → H2O + C6H3N3O6