A 0.7221 -g sample of a new compound has been analyzed and...

A 0.7221 -g sample of a new compound has been analyzed and found to contain the following masses of elements: carbon, $0.2990 \mathrm{~g} ;$ hydrogen, $0.05849 \mathrm{~g} ; $ nitrogen, $0.2318 \mathrm{~g} ;$ oxygen, $0.1328 \mathrm{~g} .$ Calculate the empirical formula of the compound.

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Key Concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms in a compound. It is derived by converting the masses of each element present in the compound to moles, then dividing by the smallest number of moles to obtain the simplest integer ratio, which forms the empirical formula.

Mole Concept

The mole concept is fundamental in chemistry for converting between the mass of a substance and the number of atoms or molecules present. By using the atomic or molar mass of each element, one can convert the measured mass to moles, which allows for a meaningful comparison between different elements in a compound.

Conversion of Mass to Moles

This concept involves dividing the mass of an element by its atomic mass to determine the number of moles present. This step is crucial in empirical formula calculations as it provides a standardized measure (moles) that reflects the actual number of atoms, independent of their masses.

Simplification to Whole-Number Ratios

After converting the masses to moles, the resulting mole amounts are divided by the smallest among them to find the simplest ratio. If the ratios are not whole numbers, they may be multiplied by an appropriate factor to convert them to whole numbers, which are then used to write the empirical formula.

Transcript

00:01 All right, so in this question, we have an unknown compound made up of these four elements, and we are trying to find the molecular formula.

00:11 Okay, so here we've been given the mass of each element within the compound.

00:23 So for carbon, it was 0 .299 grams.

00:31 Okay, this is all units of grams.

00:37 Just bracket it up top there.

00:40 Hydrogen, we've been given 0 .5849 grams.

00:56 For nitrogen, we've been given 0 .2318 grams.

01:11 And oxygen, 1 .328 grams.

01:22 Okay.

01:22 Okay, so in order for us to find the amount of each element present to write the molecule, we need to know the ratio of, like, how much of each atom is there.

01:40 So to find that ratio, we can't work with mass.

01:43 We must use moles in the unit's mole.

01:51 Okay.

01:51 So in order to find moles, we simply take mass and divide it by molar mass, right? okay.

02:06 So we know the molar mass of all of these atoms by looking at our periodic table.

02:13 So we'll just take this mass and divide it by the molar mass.

02:16 And here's what we get.

02:19 So for carbon, we take 0 .29 grams and divide it by the 12 .1 .4...