How would the calculation percent of oxalate of the unknown affected when weighting out the unknown solid the actual mass is .758 grams however the mass is mistakenly recorded as .785 grams
mols oxalate = grams/molar mass. Too high grams means too high mols.
Convert to mols KMnO4 means too high mols KMnO4.
M KMnO4 = mols/L. Too many mols means M too high.
Ok thanks. I understand
To calculate the percent of oxalate in the unknown solid, you need to know the actual mass of the unknown solid and the mass of oxalate in the unknown solid.
In this case, the actual mass of the unknown solid is recorded as 0.758 grams, but it was mistakenly recorded as 0.785 grams.
To determine the effect on the percent of oxalate calculation, you need to consider the molar mass of oxalate. Let's assume the molar mass of oxalate is X g/mol.
The percent of oxalate in the unknown solid can be calculated using the formula:
Percent of oxalate = (Mass of oxalate / Mass of unknown solid) * 100
Since the mass of oxalate is not given, let's consider it as 'Y' grams.
Given that the actual mass of the unknown solid is 0.758 grams, the equation becomes:
(Y / 0.758) * 100
Now, if the mass of the unknown solid was mistakenly recorded as 0.785 grams, the calculation would become:
(Y / 0.785) * 100
This change in the recorded mass will affect the resulting value of Y, which represents the mass of oxalate in the unknown solid. As a result, it will affect the calculated percent of oxalate.
To determine the actual effect, you would need to recalculate the percent of oxalate using the correct mass (0.758 grams) and compare it to the percent of oxalate calculated using the mistakenly recorded mass (0.785 grams). The difference between the two percentages would indicate the impact of the recording error on the percent of oxalate calculation.