A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 of the compound produced 0.213g CO2 and 0.0310 g H20. In another experiment, it is found that 0.103g of the compound produces 0.0230 g NH3. What is the empirical formula of the compound?
this is what i tried....
Compound (0.157g) + O2--> CO2(.213g) +H2O(.0310g)
help please!
I would convert 0.213g CO2 to g C, then to %C. Convert 0.0310 g H2O to grams H, then top %H. Add g H to g C and subtract from 0.157 to obtain g O, and convert to %O. Then for the separate run on N, convert g NH3 to g N and to %N.
Take a 100 g sample and that will provide you with those percentages C, H, N, O. (For example, 60% C will become 60.0 g C).
Convert g C to moles.
Convert g H to moles.
Convert g O to moles.
Convert g N to moles.
Finally, you want to find the ratio of each of these elements to each other in small whole numbers. The easy way to do that is to divide the smallest number by itself which assures of 1.000 for that number. Divide all of the other numbers by that same small number, then round to whole numbers. That should be the empirical formula. [Note: in rounding the numbers, don't round numbers that are between xx.25 and xx.75--for example 1:1.5 would become 2:3 and not 1 to 1 or 1 to 2.]
Check my thinking.
C18 H13 N5 O22?
To find the empirical formula of the compound, we need to determine the ratios of the elements present in it.
1. Start by calculating the moles of CO2 and H2O produced in the combustion reaction.
Moles of CO2 = mass of CO2 / molar mass of CO2
= 0.213 g / 44.01 g/mol (molar mass of CO2)
= 0.004843 mol
Moles of H2O = mass of H2O / molar mass of H2O
= 0.0310 g / 18.02 g/mol (molar mass of H2O)
= 0.001719 mol
2. Next, calculate the moles of carbon and hydrogen present in the compound.
Moles of C = moles of CO2 (using stoichiometry)
= 0.004843 mol
Moles of H = moles of H2O / 2 (using stoichiometry)
= 0.001719 mol / 2
= 0.0008595 mol
3. Now, let's calculate the moles of nitrogen.
Moles of N = mass of NH3 / molar mass of NH3
= 0.0230 g / 17.03 g/mol (molar mass of NH3)
= 0.001350 mol
4. Finally, calculate the mole ratios of C, H, N, and O (assuming all the other elements are oxygen).
Divide the number of moles of each element by the smallest number of moles to obtain the simplest whole-number ratio.
C: 0.004843 mol / 0.0008595 mol = 5.64
H: 0.0008595 mol / 0.0008595 mol = 1.00
N: 0.001350 mol / 0.0008595 mol = 1.57
The ratio suggests that the empirical formula of the compound is C5.64H1.00N1.57. However, we need to convert the ratios to whole numbers by multiplying all the subscripts by 2 to get whole-number values.
Multiplying by 2 gives us the empirical formula of the compound: C11H2N3.
So, the empirical formula of the compound is C11H2N3.
To find the empirical formula of the compound, we need to determine the ratio of the number of atoms of each element in the compound.
Let's start by calculating the number of moles of each compound produced during combustion:
Moles of CO2 = Mass of CO2 / Molar mass of CO2
Molar mass of CO2 = molar mass of C + 2*(molar mass of O) = 12.01 g/mol + 2*(16.00 g/mol) = 44.01 g/mol
Moles of CO2 = 0.213 g / 44.01 g/mol = 0.00484 mol
Moles of H2O = Mass of H2O / Molar mass of H2O
Molar mass of H2O = 2*(molar mass of H) + (molar mass of O) = 2*(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Moles of H2O = 0.0310 g / 18.02 g/mol = 0.00172 mol
Next, we calculate the number of moles of nitrogen and ammonia:
Moles of NH3 = Mass of NH3 / Molar mass of NH3
Molar mass of NH3 = (molar mass of N) + 3*(molar mass of H) = 14.01 g/mol + 3*(1.01 g/mol) = 17.04 g/mol
Moles of NH3 = 0.0230 g / 17.04 g/mol = 0.00135 mol
Now, we can determine the ratio of the number of moles of each element in the compound. To do this, we divide each number of moles by the smallest number of moles (in this case, 0.00135 mol):
Carbon (C) = 0.00484 mol / 0.00135 mol ≈ 3.59
Hydrogen (H) = 0.00172 mol / 0.00135 mol ≈ 1.27
Nitrogen (N) = 0.00135 mol / 0.00135 mol = 1
Oxygen (O) = (0.00484 + 0.00172) mol / 0.00135 mol ≈ 5.56
The ratio of these values can be approximated to give the empirical formula. In this case, it is C3.6H1.3N1O5.6. To simplify, we round the values to the nearest whole numbers and write the empirical formula as C3H1N1O6.
Therefore, the empirical formula of the compound is C3H1N1O6.