Ephedrine, which is used as a bronchodilator,
has the chemical formula C10H15NO. What
mass of H2O would be produced during a
combustion analysis of 0.4817 g of ephedrine?
Answer in units of g.
Here is a sample problem I have posted on the process to use in working simple stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Thank you
To find the mass of H2O produced during the combustion analysis of ephedrine, we need to calculate the number of moles of ephedrine and then determine the ratio of moles of water produced.
The molecular formula of ephedrine is C10H15NO. The molar mass of H2O is approximately 18.015 g/mol.
1. Calculate the molar mass of ephedrine:
C: 12.01 g/mol x 10 = 120.1 g/mol
H: 1.008 g/mol x 15 = 15.12 g/mol
N: 14.01 g/mol x 1 = 14.01 g/mol
O: 16.00 g/mol x 1 = 16.00 g/mol
Total molar mass of ephedrine = 120.1 + 15.12 + 14.01 + 16.00 = 165.23 g/mol
2. Calculate the number of moles of ephedrine:
moles of ephedrine = mass of ephedrine / molar mass of ephedrine
moles of ephedrine = 0.4817 g / 165.23 g/mol
moles of ephedrine ≈ 0.00292 mol
3. Determine the ratio between ephedrine and water:
From the balanced combustion equation, we know that 9 moles of water are produced for every 1 mole of ephedrine.
4. Calculate the mass of water produced:
mass of water = moles of ephedrine x (moles of water / moles of ephedrine) x molar mass of water
mass of water = 0.00292 mol x (9 mol H2O / 1 mol ephedrine) x 18.015 g/mol
mass of water ≈ 0.473 g
Therefore, during the combustion analysis of 0.4817 g of ephedrine, approximately 0.473 g of H2O would be produced.
To determine the mass of water produced during the combustion analysis of ephedrine, we need to calculate the stoichiometry of the reaction. The balanced equation for the combustion of ephedrine can be represented as:
C10H15NO + aO2 → bCO2 + cH2O + dN2
Where "a," "b," "c," and "d" are the stoichiometric coefficients. In this case, we are interested in the stoichiometric coefficient of water (c).
To find the stoichiometry, we need to know the molar mass of ephedrine (C10H15NO). By summing the molar masses of the constituent atoms, we can determine it.
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.01 g/mol
Molar mass of O = 16.00 g/mol
Therefore, the molar mass of ephedrine:
(10 × molar mass of C) + (15 × molar mass of H) + (1 × molar mass of N) + (1 × molar mass of O) = (10 × 12.01 g/mol) + (15 × 1.01 g/mol) + (1 × 14.01 g/mol) + (1 × 16.00 g/mol) = 165.23 g/mol
Now we can proceed with the stoichiometric calculation. We know that 0.4817 g of ephedrine is combusted, and we want to find the mass of water produced.
First, we need to determine the number of moles of ephedrine burned:
Moles of ephedrine = Mass of ephedrine / Molar mass = 0.4817 g / 165.23 g/mol
Next, we use the stoichiometric coefficients to relate the moles of ephedrine to the moles of water produced. From the balanced equation, we can see that the stoichiometric coefficient of water (c) is equal to 1.
Therefore, the moles of water produced = Moles of ephedrine × Stoichiometric coefficient of water (c) = Moles of ephedrine × 1
Finally, we convert the moles of water back to grams:
Mass of water = Moles of water produced × Molar mass of H2O = Moles of water produced × (2 × molar mass of H + 1 × molar mass of O) = Moles of water produced × (2 × 1.01 g/mol + 16.00 g/mol)
By plugging in the values, you can calculate the mass of water produced during the combustion analysis.