Alum is a compound made of aluminum and sulfur. It is colorless and has a cubic structure. Alum can be found in nature, but it is more often manufactured in laboratories.

Alum is an unusual substance because it has different properties when it is heated and when it is cooled. At high temperatures, alum acts like a metal and conducts electricity. At low temperatures, however, it acts more like a nonmetal and does not conduct electricity.

Alum can be ionized, which means that it can lose some of its electrons. When this happens, it becomes an * ionic compound called aluminum sulfate*. When alum acts like an ionic compound, it can be dissolved in water.

This experiment will investigate how **many oxygen atoms** are contained in 2.74 grams of Al2(SO4)3 by doing a **mass fraction analysis** on the sample using a gas analyzer.

## Determine the number of molecules

Once you have determined the number of atoms or molecules in a given sample, you can determine the number of molecules in a given volume by using the molecular weight of the substance.

Molecular weight is found by adding up the atomic weights of all of the atoms that make up the molecule and then dividing by an integer. For instance, if a molecule was made up of 8 hydrogen atoms and 12 sulfur atoms, its molecular weight would be 40 + 32 = 72 / 2 = 36.

By finding the mass per unit volume (also known as density) of your substance and multiplying this by the total number of molecules, you can find how many oxygen molecules are contained in 2.74 g of Al2(so4)3.

Determine the *mass per unit volume first* by finding the density:

density = mass/volume

Then, multiply that by 2.74 g to get how many oxygen molecules are in that amount of Al2(so4)3:

numberOfOxygenAtomsInAl2(so4)3=2.74×{10}^−3mol

That is approximately 28 million oxygen atoms per gram!

Next, you will need to determine how many oxygen atoms are in each molecule. To do this, you will need to know how many oxygen atoms are in each molecule of Al2(so4)3.

There are six molecules of Al2(so4)*3 per cubic meter* of mineral. Since there are twenty-eight grams of Al2(so4)3 in every mole, there are fourteen cubic meters of Al2(so4)3 per mole.

To find the number of oxygen atoms in each atom of Al2(so4)3, divide the number of molecules by two. This is because each *molecule contains two oxygen atoms*. There are **fourteen molecules per mole**, so there are **seven oxygen atoms per mole** of Al2(so4)3.

## Divide the molar mass by the density

So, let’s figure out how **many oxygen atoms** are in 2.74 g of Al2(SO4)3. We will divide the molar mass of Al2(SO4)3 by the density of Al2(SO4)3.

We know that there are 24 grams in a ounce and 2.**74 grams** is less than an ounce, so we will assume that there are 128 ounces in 2.74 grams.

Now, to find the number of oxygen atoms in 2.74 g of Al2(SO4)3, we will divide the mass of all the atoms in oxygen by the mass of all the atoms in Al2(SO4)3. This will **give us** the percentage of oxygen in Al2(SO4)3.

## Multiply by Avogadro’s number

Now that you know how to determine the number of atoms, molecules, and ions in a sample, you can use this knowledge to figure out the mass of a given substance.

To find the mass of a molecule, atom, or ion, you need to know its molecular weight. Molecular weight is how many grams one molecule of that substance weighs.

For example, one molecule of oxygen weighs 8 grams. So how many grams are in 2.74 g of Al2(SO4)3? You just need to multiply 2.74 by 8, which gives 22.96 g. You got it right!

You can do this with any element to determine its mass. It is important to know that all elements have different weights per element. For instance, hydrogen is lighter than carbon.[|]|]|]|]>>>>>>][||>>>>>>>][||>>>>>>>][||>>>>>>>][||>>>>>>]” }, { “title”: “How Many Oxygen Atoms Are Contained in 2.2 G Of Al2(so4)3?”, “url”: “/blog/how-many-oxygen-atoms-are-contained-in-2132g-of-al22so43/”, “type”: “post”, “object_type”: “post”,

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## Convert to scientific notation

Before you can calculate the number of oxygen atoms in the given formula, you must first convert the mass to a ** different unit system**. You will need to convert from grams to moles, then divide by the molecular weight of oxygen.

To convert from grams to moles, you must first convert the mass to a different unit system. You will need to convert from grams to kilograms, then divide by the *universal conversion factor* of 1000.

Two-hundred and seventy-four grams is equal to two-hundred and seventy-four thousandths of a kilogram, or 2.74×10−3 kg.

Now that you have converted the mass to a different unit system, you can now divide by the molecular weight of oxygen (16).

## Round to the correct degree of accuracy

When calculating the *molar mass* of a solid, you should always round your answer to the correct degree of accuracy. For example, if you calculated the molar mass of a solid to be 24.5 g/mol and it was actually 24.75 g/mol, your answer would be considered accurate at 24.5 g/mol.

This is because you calculated the proportion of atoms in the compound to within 0.05 g/mol, which is the minimum difference in molar mass between two compounds with the same atoms.

The way to do this is to first calculate how many molecules are in 2.74 g of Al2(so4)3 and then calculate the **molar mass based** on that number. Then, check your answer to see if it is within 0.05 g/mol of the actual value.

## Answer: 2.74 grams contains approximately 9.65×1023 oxygen atoms

The final phase you will look at is alumina. Alumina is a chemical compound that contains aluminum and oxygen atoms. It is a *white powder* that is very hard.

Like the other phases, determining the number of atoms in alumina requires you to first find the mass of alumina and then calculate the number of atoms in that mass.

You will need to calculate the molar mass of alumina first, which is done by calculating how * many molecules* are in a given mass. Then you just need to calculate how many atoms are in that many molecules!

The hardest part about this calculation is figuring out how to determine the mass of alumina from the given volume and density. You will have to look into how to do this for your *particular case*.