A compound consists only of carbon, hydrogen, nitrogen, and oxygen.  Combustion of 10.68 g of the compound  produced 16.01 g carbon dioxide and 4.37 g of water.  the molsr mass of the compound is 176.1 g/mol. what are the empiricaland molecular formulas of the compound?

To determine the empirical and molecular formulas of the compound, we need to follow a step-by-step process.

Step 1: Calculate the moles of carbon dioxide (CO2) produced.
The molar mass of carbon dioxide is 44.01 g/mol.
Moles of CO2 = mass of CO2 / molar mass of CO2
Moles of CO2 = 16.01 g / 44.01 g/mol
Moles of CO2 = 0.363 mol

Step 2: Calculate the moles of water (H2O) produced.
The molar mass of water is 18.015 g/mol.
Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 4.37 g / 18.015 g/mol
Moles of H2O = 0.243 mol

Step 3: Calculate the moles of carbon, hydrogen, and oxygen in the original compound.
Moles of C = Moles of CO2
Moles of C = 0.363 mol

Moles of H = (Moles of H2O) × 2 (since there are two moles of hydrogen in one mole of water)
Moles of H = 0.243 mol × 2 = 0.486 mol

Moles of O = (Moles of CO2) × 2 (since there are two moles of oxygen in one mole of carbon dioxide)
Moles of O = 0.363 mol × 2 = 0.726 mol

Step 4: Determine the simplest ratio of the moles of carbon, hydrogen, and oxygen.
Divide the number of moles by the smallest number of moles.
C: 0.363 mol / 0.243 mol = 1.496
H: 0.486 mol / 0.243 mol = 2.000
O: 0.726 mol / 0.243 mol = 2.988

To simplify, we multiply each ratio by the inverse of the smallest value (2.000).
C: 1.496 × 2 = 2.992
H: 2.000 × 2 = 4.000
O: 2.988 × 2 = 5.976

Step 5: Determine the empirical formula.
The empirical formula represents the simplest whole-number ratio of the atoms in a compound. Since we cannot have fractional atoms, it is best to round all values to the nearest whole number.
C: 3
H: 4
O: 6

So, the empirical formula of the compound is C3H4O6.

Step 6: Calculate the empirical formula's molar mass.
To calculate the empirical formula's molar mass, we need to sum up the molar masses of each atom in the formula.
Molar mass of carbon (C) = 12.01 g/mol
Molar mass of hydrogen (H) = 1.008 g/mol
Molar mass of oxygen (O) = 16.00 g/mol

Empirical formula's molar mass = (3 × molar mass of C) + (4 × molar mass of H) + (6 × molar mass of O)
Empirical formula's molar mass = (3 × 12.01 g/mol) + (4 × 1.008 g/mol) + (6 × 16.00 g/mol)
Empirical formula's molar mass = 36.03 g/mol + 4.032 g/mol + 96.00 g/mol
Empirical formula's molar mass = 136.062 g/mol

Step 7: Calculate the molecular formula.
The molecular formula represents the actual number of atoms of each element in a molecule.
Molar mass of the compound = 176.1 g/mol (given)

To determine the molecular formula from the empirical formula, divide the molar mass of the compound by the molar mass of the empirical formula.
Molecular formula = (Molar mass of the compound) / (Empirical formula's molar mass)
Molecular formula = 176.1 g/mol / 136.062 g/mol
Molecular formula ≈ 1.294

Since we can't have fractional molecular formulas, we will round the value to the nearest whole number.
Approximately, the molecular formula is 1.

Therefore, the empirical formula of the compound is C3H4O6, and the molecular formula is 1.