Stoichiometry and Analytical Chemistry - MCAT Physical

Any combustion reaction of a hydrocarbon involves oxygen gas as a reactant, and produces carbon dioxide and water as products. In this case, two moles of hexane react with nineteen moles of oxygen to produce twelve moles of carbon dioxide and fourteen moles of water. Carbon dioxide is a gas at room temperature, and water is a liquid.

Hexane:

Reaction:

Now we can begin to balance the reaction.

Everything is balanced except oxygen; there is an odd number of oxygen to the right and an even number to the left. We can adjust this by multiplying everything by two.

First convert grams of ammonium sulfate to moles, then use the mole-to-mole ratio between nitrogen and ammonium sulfate. Finally, convert moles of nitrogen back into grams.

Given that the molar mass of oxygen is about 16g, and molar mass of manganese is about 55g, contains 165g of manganese and 64g of oxygen, for a total of 229g.

165g/229g = 0.72

So, the ratio of manganese to oxygen in this compound is 72% manganese by mass.

To find the molecular weight (mass) of a molecule, simply add up the atomic weights of each atom within the molecule. The units used will be amu (atomic mass units) for molecular weight and g/mol for molar mass.

First find the molar mass of water (H2O). You should be comfortable with the molar masses of hydrigen and oxygen from memory to reduce time on the MCAT exam.

Next, solve for moles.

It is important to remember that one mole of an element contains 6.022 * 1023 atoms. Since there are two moles of carbon dioxide (CO2), we can conclude that there are two moles of carbon. As a result, there are atoms.

When finding the empirical formula for a compound, it helps to imagine a 100g sample of the molecule. This way, the percentages of the atoms can be converted to amounts in grams. At this point, dividing each amount by the atom's molar mass results in the values that will be compared to one another in order to find the ratio of atoms in the molecule.

(carbon)

(hydrogen)

(oxygen)

(nitrogen)

From these calculations, we can set up a ratio.

3.45C : 6.9H : 1.72O : 1.72N

If we divide this ratio by the smallest value (1.72), we can see that it can be reduced.

2C : 4H : 1O : 1N

As a result, the empirical formula for the compound is C2H4ON.

In order to find the mass percentage of an atom in a molecule, start by finding the total mass of one mole of the molecule. Glucose has 180 grams per mol.

Next, we determine the mass of the carbon atoms in one mole of the molecule. One carbon mole has a mass of 12 grams. Multiplied by the six carbon atoms in glucose gives a mass of 72 grams.

Finally, we divide the mass of carbon by the mass of the molecule.

So, 40% of glucose's mass is made up of carbon.

The atomic mass of an element is determined by the proportional mass of each elemental isotope. We know that there are only two isotopes of lithium; therefore, their percentages must add to 100%.

The atomic mass will be equal to the mass of each isotope multiplied by its abundance.

We can substitute an algebraic expression to solve for one of our variables.

Using this value, we can solve for the abundance of the other isotope.

Converting these values to percentages gives us our final answer.

Aluminum oxide has the formula .

Aluminum has a molecular weight of , and oxygen has a weight of . Using these values, we can calculate the molecular weight of aluminum oxide.

The mass percentage is given by the mass of aluminum divided by the total molecular weight.

Sodium sulfate is given by the formula:

To find the percentage by weight, we will need to divide the mass of sodium in the molecule by the total molecular mass.

Convert the ratio to a percentage.

A full liter of a one molar solution of sodium hydroxide would contain one mole of sodium ions, or ions. Here, you have only one tenth the volume, so multiply the number in one mole by one tenth.

Now that we have reduced the volume, we need to account for the concentration.

4-octene looks like this:

To get an empirical formula, we find the ratio of each element within the compound and make it as low as possible. We have eight carbons and sixteen hydrogens. The ratio of carbons to hydrogens is 8-to-16, which reduces to 1-to-2. The full formula for 4-octene is , and the empirical formula is .

An empirical formula must be written as the most simplified ratio of the elements that the compound contains. For example, is empirical because it cannot be simplified any further; the ratio of its atoms is 1:1:4.

The formula for glucose, , can be simplified by a factor of six. The empirical formula for glucose would be .

The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.

The molecular weight of carbon is , hydrogen is , and oxygen is .

The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.

After finding the moles of each element, you have to find the smallest whole number ratio of each element. The smallest whole number ratio can be found by dividing moles of each element by the lowest mole quantity (in this case, of oxygen).

You are left with carbons, hydrogens, and oxygen. The empirical formula for this compound is .

The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.

After finding the moles of each element, you have to find the smallest whole number ratio of each element. The smallest whole number ratio can be found by dividing moles of each element by the lowest mole quantity (in this case, of oxygen). You are left with carbons, hydrogens, and oxygen. The empirical formula for this compound is .

To find the molecular formula of the compound you need to divide the molecular weight of the actual compound by the molecular weight of the empirical formula. The molecular weight of the empirical formula is:

Dividing the molecular weight of the actual compound () by the molecular weight of empirical formula gives:

This means that the empirical formula must be multiplied by three to get the molecular formula; therefore, the molecular formula is . Compared to the empirical formula, the molecular formula contains more carbon atoms and more oxygen atoms.

Remember that the empirical formula relies on the ratio of the moles of elements. To get the number of atoms, you would have to multiply the moles of each element by the Avogadro’s number (). You would use this number for every element (Avogadro’s number doesn’t change for each element). This means that the ratio of the number of atoms will be the same as the ratio of the number of moles, and you will get the same empirical formula. There is a constant relationship between the number of moles and the number of atoms in a sample.

A mole is defined as the number of atoms present in of carbon-12. Obtaining a sample of carbon-12 will give you of carbon-12. You can also find the number of moles by multiplying the mass by the molecular weight of carbon:

Remember that a mole of any element contains atoms (Avogadro’s number). Since we have exactly one mole in the sample, there will be exactly atoms of carbon-12.

To find the empirical formula, use the mass percentage of each element to find mole ratios based on a hypothetical sample of .

We see that there is a 1:1 mole ratio for carbon to hydrogen, making the empirical formula .

The next step will be to find molecular formula by dividing the molar mass of the compound by the molar mass of the empirical formula.

Multiply the subscripts of the empirical formula for each element by six to get the molecular formula: .

An empirical formula is the simplest form of a molecular formula that still retains the ratio of the elements. If a formula can be divided by a whole number, it is a molecular formula and not an empirical formula. A molecular formula is the exact identity of a compound, showing the total number of atoms used to create the compound.

The only given answer that is not divisible by a whole number is , making it an empirical formula. It is also the molecular formula for both acetaldehyde and ethanol, depending on molecular geometry and orientation.