Molecular Weight, Molecular Formula, and Moles - MCAT Chemical and Physical Foundations of Biological Systems

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.

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: .

The molecular formula is the same as the empirical formula if it cannot be reduced by any whole number. Any formula containing a single atom of any given element must be an empirical formula as well. The formula for DEET is . Since this contains a single atom each of oxygen and nitrogen, it cannot be further reduced and must be an empirical formula as well.

The molecular formula for ribose is , which can be reduced by a factor of five. The empirical formula for ribose (and most other carbohydrates) is .

The other two options require us to calculate pass percentages based on the given molecular formulas.

Although we can look at the formulas for chlorophyll and ethyl butyrate to deduce that oxygen makes up a larger percentage of the latter, we can double check mathematically. In order to find which compound contains oxygen in a larger percentage, divide the molar mass of the oxygen in the compound by the molar mass of the entire compound.

For ethyl butyrate:

For chlorophyll:

We see that it is in fact true that there is a larger percentage of oxygen in ethyl buyrate.

Next, find out if hydrogen and carbon make up the largest mass percentage of ribose by using the same method:

This amounts to 46.7% of the molecular mass, meaning that oxygen must account for the remaining 53.3%. Oxygen thus makes up a greater mass percentage of ribose than hydrogen and carbon combined.

Given the percentages by mass of the compound, we can convert the percentages to grams by considering a 100-gram sample of the compound. The next step involves dividing each mass by the element's molar mass.

By dividing each value by the smallest molar quantity, we can determine the ratio of elements in the compound.

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

The empirical formula for carbohydrates is , where is the number of carbon atoms. The question states that molecule A has 6 carbons; therefore, and the molecular formula for molecule A is . The molecular weight (MW) of this compound is calculated using the molecular weight of each atom. The MW of each atom is obtained from the periodic table and is multiplied by the number atoms.

So, the molecular weight of molecule A is . Since it is also a carbohydrate, molecule B will have the same empirical formula as molecule A. Molecule B has half the molecular weight of molecule A; therefore, molecule B must have half the atoms as molecule A. The molecular formula of molecule B is . Note that MW of is .

The MW contribution of oxygen to each compound is calculated as follows.

%MW of oxygen for molecule A =

%MW of oxygen for molecule B =

Therefore, oxygen contributes to 53% of MW in both compounds.

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.

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.

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: .

Given the percentages by mass of the compound, we can convert the percentages to grams by considering a 100-gram sample of the compound. The next step involves dividing each mass by the element's molar mass.

By dividing each value by the smallest molar quantity, we can determine the ratio of elements in the compound.

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

The empirical formula for carbohydrates is , where is the number of carbon atoms. The question states that molecule A has 6 carbons; therefore, and the molecular formula for molecule A is . The molecular weight (MW) of this compound is calculated using the molecular weight of each atom. The MW of each atom is obtained from the periodic table and is multiplied by the number atoms.

So, the molecular weight of molecule A is . Since it is also a carbohydrate, molecule B will have the same empirical formula as molecule A. Molecule B has half the molecular weight of molecule A; therefore, molecule B must have half the atoms as molecule A. The molecular formula of molecule B is . Note that MW of is .

The MW contribution of oxygen to each compound is calculated as follows.

%MW of oxygen for molecule A =

%MW of oxygen for molecule B =

Therefore, oxygen contributes to 53% of MW in both compounds.

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.

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.

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 .

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.

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.

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.