if a student finds the molecular weight of an unknown liquid to be 115 g/mole and the molecular formula of the liquid is CHCl3, what is the percent error in the student's molecular weight?
%error = (115-119)/119]*100 = ?
To calculate the percent error in the student's molecular weight, we need to compare it to the actual molecular weight of CHCl3.
The molecular weight of CHCl3 can be calculated as follows:
1 carbon atom (C) = 12.01 g/mol
1 hydrogen atom (H) = 1.01 g/mol
3 chlorine atoms (Cl) = 35.45 g/mol each
Molecular weight of CHCl3 = 12.01 g/mol + (1.01 g/mol × 1) + (35.45 g/mol × 3)
= 12.01 g/mol + 1.01 g/mol + 106.35 g/mol
= 119.37 g/mol
Now, we can calculate the percent error using the following formula:
Percent error = ((Measured value - Actual value) / Actual value) × 100
Percent error = ((115 g/mol - 119.37 g/mol) / 119.37 g/mol) × 100
= (-4.37 g/mol / 119.37 g/mol) × 100
= -0.0365 × 100
= -3.65%
The percent error in the student's molecular weight is approximately -3.65%. Note that the negative sign indicates an underestimate of the molecular weight.
To calculate the percent error in the student's molecular weight, we need to compare their result with the correct value.
The correct molecular weight of CHCl3 can be obtained by adding up the atomic masses of the atoms in the molecule. Let's calculate it:
Carbon (C): Atomic weight = 12.01 g/mole
Hydrogen (H): Atomic weight = 1.008 g/mole
Chlorine (Cl): Atomic weight = 35.45 g/mole
Molecular weight of CHCl3 = (12.01 * 1) + (1.008 * 1) + (35.45 * 3) = 50.49 g/mole
Now, we can calculate the percent error using the following formula:
Percent error = [(|Correct value - Experimental value|) / Correct value] * 100
Percent error = [(|50.49 - 115|) / 50.49] * 100
Percent error = (64.51 / 50.49) * 100
Percent error ≈ 127.70%
Therefore, the student's percent error in calculating the molecular weight of the unknown liquid is approximately 127.70%.