What are the empirical formulas of each of the following...

Name: Problem 103, Chapter 3: Chemistry
Uploaded: 2020-06-09T04:15:22-08:00
Duration: 5 min 27 s
Channel: Nadia Lara
Description: What are the empirical formulas of each of the following substances? (a) Ibuprofen, a headache remedy: 75.69% C, 15.51% O, 8.80% H (b) Magnetite, a naturally occurring magnetic mineral: 72.36% Fe, 27.64% O (c) Zircon, a mineral from which cubic zirconia is made: 34.91% O, 15.32% Si, 49.77% Zr

What are the empirical formulas of each of the following substances? (a) Ibuprofen, a headache remedy: 75.69% C, 15.51% O, 8.80% H (b) Magnetite, a naturally occurring magnetic mineral: 72.36% Fe, 27.64% O (c) Zircon, a mineral from which cubic zirconia is made: 34.91% O, 15.32% Si, 49.77% Zr

What are the empirical formulas of each of the following substances?
(a) Ibuprofen, a headache remedy: 75.69% C, 15.51% O, 8.80% H
(b) Magnetite, a naturally occurring magnetic mineral: 72.36% Fe, 27.64% O
(c) Zircon, a mineral from which cubic zirconia is made: 34.91% O, 15.32% Si, 49.77% Zr

Key Concepts

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element present in the substance. It is derived by simplifying the actual numbers of atoms, which may be represented in a molecular formula, to the smallest possible integers that maintain the same proportion of elements.

Percent Composition

Percent composition refers to the percentage by mass of each element in a compound. It provides a way to understand the relative amounts of each element that make up a substance and is typically used as the starting point in determining the empirical formula.

Conversion to Moles

Converting percent composition to moles involves dividing the percentage (usually taken as grams per 100 grams of compound) of each element by its atomic mass. This step translates the mass percentages into quantitative measures which can then be compared to determine ratios between the elements.

Mole Ratio and Simplification

After calculating the moles of each element, the next step is to obtain the simplest whole-number ratio. This is done by dividing each element's mole value by the smallest of the mole values obtained, and, if necessary, multiplying the results by an appropriate factor to achieve whole numbers.

Transcript

00:01 Start by assuming that these percentages are of a 100 grand sample.

00:06 That means that we can change these units to grams and then divide through by the molar masses of each.

00:20 These i get too many zeros from the periodic table like so.

00:31 Dividing through for each one gives me 6 .302 moles of carbon, 0 .969.

00:43 94 moles of oxygen and 8 .713 moles of hydrogen.

00:51 I'm going to write these as a formula.

00:55 So this is c6302 0 .9695, h8 .713.

01:07 Remember that for formulas we need to have whole number ratios.

01:11 To get those whole number ratios, i'm going to divide through each of these subscripts by the smallest one.

01:20 That's the 9695 that goes with oxygen.

01:30 Simplifying gives carbon 6 .50, oxygen 1 .00 and hydrogen 8 .98.

01:42 Now the hydrogen is close enough to nine, like within two hundreds, that we could reasonably round that up to 9 .0 to just the 9.

01:52 But the 6 .50 is troubling.

01:55 We cannot round up.

01:56 That is not close enough to a whole number.

01:59 Instead, what we're going to do is we're going to double the formula.

02:07 And that's going to give us carbon 13 -01h18, which actually i should probably write as it is done more commonly with carbon first, hydrogen 18, and oxygen.

02:24 Forgive me, 1 times 2 should be 2 like so.

02:34 So this is my empirical formula for part a.

02:39 In part b, we're going to do the same thing...

Please give Ace some feedback