Calculate Percentage Purity by Mass of C10H7OH Contaminated with LiBr

[physicsss]

A sample of C10H7OH was found to be contaminated with a small amount of LiBr. The combustion analysis of the sample gave the following results: 77.48% C and 5.20% H. Assume that the only species present are C10H7OH and LiBr, calculate the percentage purity by mass of the sample.

First of all, what is the percentage of purity by mass...? And second, does it mean that 100%-77.48%-5.20% gives the percentage of O and LiBr?

 

[Gokul43201]

First of all, what is the percentage of purity by mass...? And second, does it mean that 100%-77.48%-5.20% gives the percentage of O and LiBr?

If you had 100g of the sample and found that it contained x grams of LiBr, then the percentage purity would be 100-x.

From the formula of the compound, can you not determine how much O there is in it ?

 

[ravenprp]

Percentage purity by mass refers to the amount of the desired substance (in this case, C10H7OH) present in a sample compared to the total mass of the sample. This calculation is important because it gives an indication of the quality and effectiveness of a substance.

To calculate the percentage purity by mass of the sample contaminated with LiBr, we need to first determine the amount of C10H7OH present in the sample. We can do this by using the given percentages of carbon and hydrogen, assuming that all the carbon and hydrogen present in the sample come from C10H7OH.

To find the amount of C10H7OH, we can use the molar mass of the compound (C10H7OH = 159.17 g/mol) and the given percentages:

Mass of C in sample = 77.48% x 159.17 g/mol = 123.33 g
Mass of H in sample = 5.20% x 159.17 g/mol = 8.28 g

Therefore, the mass of C10H7OH in the sample is 123.33 + 8.28 = 131.61 g.

Next, we need to find the total mass of the sample. Since the only other species present is LiBr, we can assume that the remaining mass is due to LiBr.

Mass of LiBr in sample = Total mass of sample - Mass of C10H7OH
= 100 g - 131.61 g = -31.61 g

However, we know that this is not possible, as the mass of LiBr cannot be negative. This suggests that there is no LiBr present in the sample and that the combustion analysis results were likely contaminated.

Therefore, the percentage purity by mass of the sample is 100% (since there is no other substance present). This means that the sample is 100% C10H7OH.

In conclusion, the sample of C10H7OH is 100% pure, as there is no evidence of LiBr contamination in the combustion analysis results. However, if there were indeed LiBr present in the sample, we would need to recalculate the percentage purity by mass taking into account the mass of LiBr.