Convert the following percent compositions into molecular...

Convert the following percent compositions into molecular formulas:
(a) Diborane: $\mathrm{H} 21.86 \%, \mathrm{B} 78.14 \% ;$ Molec. mass $=27.7 \mathrm{amu}$
(b) Trioxan: $\mathrm{C} 40.00 \%, \mathrm{H} 6.71 \%, \mathrm{O} 53.28 \% ;$ Molec. mass $=90.08 \mathrm{amu}$

Key Concepts

Percent Composition

Percent composition shows the mass percentage of each element within a compound. It provides a way to understand the relative amounts of each element in a sample, serving as the basis for determining the formula of the compound by relating the mass percentages to moles.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms of each element present in a compound. It is derived by converting the mass percentages into moles and then dividing by the smallest number of moles to obtain a ratio.

Molecular Formula

The molecular formula indicates the actual number of atoms of each element in a molecule and may be a multiple of the empirical formula. It is determined using the compound's molecular (or molar) mass in conjunction with the empirical formula.

Converting Percent Composition to Moles

This key concept involves converting the mass percentages of elements into moles by dividing each percentage by the corresponding atomic mass. This conversion is essential for determining the empirical formula by providing mole ratios for the elements present.

Transcript

00:01 To find the molecular formula, we're going to follow the following procedure.

00:06 So first, we're assuming that we have 100 grams of diborane.

00:10 And if we have 100 grams by diborin, this just makes the problem easier.

00:14 Then we can stop dealing with the percent and instead deal with grams of each of these.

00:21 We're going to find the moles of each element.

00:25 So for hydrogen, we're going to divide by the molar mass of hydrogen.

00:32 008 grams of hydrogen.

00:35 Similarly for boron, we're going to divide by its molar mass, which is 10 .81 grams of boron and every one mole of boron.

00:49 This gives us 21 .64 mole of hydrogen and 7 .228 moles of boron.

01:02 We're going to write these multiple ratios as a formula, so b7 .228h21 .64, and then recall that in our formulas, we can only have the simplest whole number ratio.

01:25 To get that simplest whole number ratio, we're going to divide through by the smaller of the two numbers.

01:33 And this gives us a formula that is b2 .9, sorry, b1 .00 h2 .99, which rounds to three.

01:49 So when we simplify this, this is equal to bh3.

01:56 That is our empirical formula that we have found.

01:59 To find the molecular formula, we need to use this mass that we were given.

02:08 So first, vh3 has the molecular mass of 13 .83 amus.

02:26 I find this using my periodic table, one times the mass of boron, three times the molar mass of hydrogen, gives us 13.

02:36 And then we see that this value is approximately half, or rather, the diburine mass is about twice as great as that...

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