How do you determine a empirical formula?

Answer 1

The empirical formula is when you can not simplify the formula any further.

Let's use the formula for glucose, C6 H12 O6

That is the molecular formula of glucose.

The Empirical Formula of Glucose would be C1 H2 O1, because you can divide each element by 6.

As for a compound such as ammonia N H3, that is it's Molecular Formula.

It's empirical formula would be N H3 as well because it can not be simplified any further.

Answer 2

You can find the empirical formula of a compound using percent composition data. If you know the total molar mass of the compound, the molecular formula usually can be determined as well. The easiest way to find the formula is:

1. Assume you have 100 g of the substance (makes the math easier because everything is a straight percent).
2. Consider the amounts you are given as being in units of grams.
3. Convert the grams to moles for each element.
4. Find the smallest whole number ratio of moles for each element

Answer 3

Example Question: What is the empirical formula for a compound if a 2.50 g sample contains .900 g of calcium and 1.60 g of chlorine?

First you have to convert the masses to moles:

.900 g Ca ÷ 40.1 g = .0224 moles Ca

1.6 g Cl ÷ 35.5 g = .0451 moles Cl

This is a ratio of .0224 to .0451. But it is unrecognizable as a small whole number ratio (see Law of Definite Proportions). A "trick": divide each number by the smallest number. This will give you a ratio that looks better and will always have a 1 as the smallest number (since you divided it by itself).

Ca = .0224 ÷ .0224 = 1

Cl = .0451 ÷ .0224 = 2.01

Since the last digit is always uncertain, we can say that the 2.01 is actually 2

So the empirical formula is: CaCl2

Example Question: What is the empirical formula of a compound with the following percentages: 40.0% Carbon, 6.71% Hydrogen, 53.5% Oxygen?

The key to this problem is to think "if I had 100 grams of the compound". If you have 100 grams of the compound, then you can use the percentages as measurements. So completing the sentence, "if I had 100 grams of the compound, then I would have 40.0 grams of hydrogen, 6.71 grams of carbon, and 53.5 grams of oxygen".

The next step is the same as in the first problem: Convert to moles

40.0 g ÷ 12.0 g = 3.33 moles Carbon

6.71 g ÷ 1.0 g = 6.7 moles Hydrogen

53.5 g ÷ 16.0 g = 3.34 moles Oxygen

Then you divide by the smallest number like in the first problem in order to determine the small whole number ratio.

3.33 ÷ 3.33 = 1 Carbon

6.7 ÷ 3.33 = 2 Hydrogen

3.34 ÷ 3.33 = 1 Oxygen

So the empirical formula is: CH2O

In general we should get ratios of atoms which are accurate to ± .02

If your numbers are not ± .02 to whole numbers or the values below, then you did something wrong!

2.02 ≈ 2

4.01 ≈ 4

3.98 ≈ 4

But there are special cases (all are ± .02)

If you have a ratio that is halfway between two numbers (1.5, 2.5, 3.5), then you multiply everything by 2

1.5:1 = 3:2

When you have a ratio that is in one-thirds or two-thirds increments (1.33, 2.33, 1.66, 2.66), then multiply everything by 3

1.33:2 = 4:6 = 2:3

1.65:1 = 5:3

When you have a ratio that is one-quarter or three-quarter increments (1.25, 1.75), then multiply everything by 4

1.26:1 = 5:4

1.74:2 = 7:8

Answer 4

the empirical formula is the simplest whole number ratio of the elements present in one molecule or formula unit of a compounds we calculate the empirical formula using the following steps:

1. note the mass of each element correctly

2. divide the atomic masses by the masses deduced in step 1

3. divide the step 2 calculation by the lowest figure

Answer 5

empirical formula,.

number of grams of an element will be given. divide that by the Atomic Mass number.

find the ratio of all the elements in the given question and find the ratio of the elements

eg: a compound of xenon contains 9.825g of xenon ; 1.200 g of oxygen ; 5.700g of fluorine

atomic mass of xenon is 131 and oxygen is 16

fluorine is 19

so, xenon : oxygen : fluorine

9.85/131 : 1.200/16 : 5.700/19

0.0752 : 0.075 : 0.3

1 : 1 : 4

so, (XeOF4)n

Rishabh Arora

Answer 6

Experimental analysis will give the elements and their percentages in the compound.

Then it becomes a math problem.

Step 1. For each element divide each percentage by the respective atomic mass. Step 2. Look at those answers. Use the smallest to divide into all of the decimal answers from step 1.

Step 3. IF all the answers are whole numbers, use them as subscripts and write the emp. formula. If they are very close (x.99 or x.98) round to whole numbers, but if the decimals are not close then think of them as fractions. Ex. .5 or anything close like .51 or .49 becomes 1/2; .33 becomes 1/3, .25 becomes 1/4 etc. Now Take the reciprocal of the fraction and multiply ALL decimal answers fromstep 2 by that whole number. ---- this creates small whole numbers to use as subscripts, write the formula.

Answer 7

The empirical formula is a bit like the lowest common multiple in math. Let us take C6H1206 which is glucose. By inspection it can be seen all the subscripts can be divided by 6 to yield CH2O and that is the empirical formula. If we took the hydrocarbon C6H12 which is cyclohexane, the empirical formula would be CH2 as the 6 and 12 subscripts can be divided by 6 to yield 1 and 2. etc.

Answer 8

CH, 58 g/mol; CH, 78 g/mol; and HgCl, 236.1 g/mol.

Answer 9

Empirical formulas are the result of chemical analysis of a compound.