A compound made up of C, H, and Cl contains 55.0 percent Cl by mass. If 9.00 g of the compound

A compound made up of C, H, and Cl contains 55.0 percent Cl...

Key Concepts

Empirical Formula Determination

Empirical formula determination involves calculating the simplest whole-number ratio of atoms present in a compound. This process typically begins by converting the mass data (often given as percentages or measured masses) into moles for each constituent element, and then simplifying the mole ratio to the smallest whole numbers.

Percent Composition

Percent composition describes the mass percentage of each element in a compound. It allows one to relate the mass of each element to the total mass of the compound, which is crucial for determining the amount of each element present when calculating the empirical formula.

Mole Concept

The mole concept is fundamental in chemistry as it provides a bridge between the atomic scale and macroscopic measurements. It relates a measurable amount of substance to a specific number of atoms or molecules, enabling conversions between mass and number of particles.

Avogadro's Number

Avogadro's number, approximately 6.022 x 10^23, is the number of atoms or molecules in one mole of any substance. It is used to convert between moles and individual particles (atoms or molecules), which is essential when determining composition from counts of atoms.

Stoichiometric Calculations

Stoichiometric calculations involve using the mole ratios derived from the empirical formula or balanced equations to compute the quantities of elements in a compound. This method is critical when converting mass percentages and counts of atoms into a standardized, simplified formula representation.

Transcript

00:01 So for this problem, we want to find the empirical formula of a compound, given that we know it is 55 % chlorine by mass, and that 9 grams of the entire compound contains 4 .19 times 10 to the 23 hydrogen atoms.

00:21 And we know that this compound contains only carbon, hydrogen, and chlorine.

00:26 So, as we always do for empirical formula problems, we're going to assume, that the compound is 100 grams.

00:36 This obviously isn't true, but because we do know a percent mass of the entire compound, this is a very easy number.

00:43 Because if we assume that the entire compound is 100 grams, that means that 55 grams of that will be chlorine.

00:56 We can easily convert this into moles of chlorine because we know that per every one mole of chlorine, that its mass is 35 .45 grams.

01:06 So that will yield 1 .1 .4 .5 grams.

01:07 So that will yield 1 .4 0 .55 moles of chlorine.

01:11 So we know that if hypothetically our compound weighed 100 grams, it would contain 1 .55 moles of chlorine.

01:18 Now we want to do the exact same thing with our hydrogen.

01:22 So we know that once again we're going to assume that our compound is 100 grams.

01:32 And we know that per every 9 grams of this compound, as you see right here, for every 9 grams of this compound, that 4 .19 times 10 to the 23, hydrogen atoms are present.

01:49 And now we want to convert this into moles of hydrogen.

01:54 We just divide by abogadjo's number to get 7 .73 moles of hydrogen.

02:06 So now that we have the moles of both chlorine and hydrogen, we want to find now the moles of carbon.

02:13 Well, we can do that by finding the masses of everything that we need.

02:19 So we already know the mass of chlorine to be 55 grams, because that's the assumption that we made at the very beginning.

02:25 So now we want to find the mass of hydrogen, so we can subtract both those values out from the total value to get the remaining mass, which is just carbon.

02:33 So we know that we're dealing with 7 .73 moles of hydrogen, and we know that in every mole of hydrogen, mass is 1 .08 grams...

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