Hello everyone!

I am currently in my fourth and final year of a **masters degree in chemistry **at the University of Sheffield.

I am here to help you with **any problems **you may have in **chemistry**, at any levels. University friends often come to me as I have a good ability of **taking a complicated principle and turning it into an easy one to understand and follow. **

I am enthusiastic about science in general and more than happy to share my knowledge with you to **relieve any stresses** you have about particular **areas or topics in chemistry**.

I look forward to helping you guys out!

Hello everyone!

I am currently in my fourth and final year of a **masters degree in chemistry **at the University of Sheffield.

I am here to help you with **any problems **you may have in **chemistry**, at any levels. University friends often come to me as I have a good ability of **taking a complicated principle and turning it into an easy one to understand and follow. **

I am enthusiastic about science in general and more than happy to share my knowledge with you to **relieve any stresses** you have about particular **areas or topics in chemistry**.

I look forward to helping you guys out!

No DBS Check

The first step here is to determine the mass of C in CO_{2} and the mass of H in H_{2}O. This is done by dividing the relative atomic mass, Mr, by the relative molecular mass of the compound, M_{CO2} or M_{H2O}, and then multiplying by the mass of compound produced.

For C:

(12.011 g / 44.009 g) x 0.3664 g = 0.1000 g

For H:

(1.0079 x 2 / 18.0148) x 0.1500 g = 0.0168 g

From here we can determine the mass of oxygen in the hydrocarbon by subtraction of the two above masses:

O: 0.2500 g - 0.1000 g - 0.0168 g = 0.1332 g

From here we determine the number of moles of each element, this is done by dividing by the mass of each element by its atomic mass:

C: 0.1000 g / 12.011 g mol^{-1 }= 0.0083 mol

H: 0.01678 g / 1.0079 g mol^{-1} = 0.0166 mol

O: 0.1332 g / 15.999 g mol^{-1} = 0.0083 mol

The final step is to divide by the smallest number that obtained, in this case that is the number of moles of C and O:

C: 0.0083 mol / 0.0083 mol = 1

H: 0.0166 mol /0.0083 mol = 2

O: 0.0083 mol / 0.0083 mol = 1

We have now arrived at a ratio of 1:2:1 for C:H:O, thus we have an empirical formula of CH_{2}O.

The first step here is to determine the mass of C in CO_{2} and the mass of H in H_{2}O. This is done by dividing the relative atomic mass, Mr, by the relative molecular mass of the compound, M_{CO2} or M_{H2O}, and then multiplying by the mass of compound produced.

For C:

(12.011 g / 44.009 g) x 0.3664 g = 0.1000 g

For H:

(1.0079 x 2 / 18.0148) x 0.1500 g = 0.0168 g

From here we can determine the mass of oxygen in the hydrocarbon by subtraction of the two above masses:

O: 0.2500 g - 0.1000 g - 0.0168 g = 0.1332 g

From here we determine the number of moles of each element, this is done by dividing by the mass of each element by its atomic mass:

C: 0.1000 g / 12.011 g mol^{-1 }= 0.0083 mol

H: 0.01678 g / 1.0079 g mol^{-1} = 0.0166 mol

O: 0.1332 g / 15.999 g mol^{-1} = 0.0083 mol

The final step is to divide by the smallest number that obtained, in this case that is the number of moles of C and O:

C: 0.0083 mol / 0.0083 mol = 1

H: 0.0166 mol /0.0083 mol = 2

O: 0.0083 mol / 0.0083 mol = 1

We have now arrived at a ratio of 1:2:1 for C:H:O, thus we have an empirical formula of CH_{2}O.

The first step here is to determine the mass of C in CO_{2} and the mass of H in H_{2}O. This is done by dividing the atomic mass by the molecular mass and then multiplying by the mass of compound produced.

For C:

(12.011 g / 44.009 g) x 4.40 g = 1.1999 g

For H:

(1.0079 x 2 / 18.0148 g) x 2.70 g = 0.3021 g

The next step is to convert these masses into moles. This is done by dividing the mass by the relative atomic mass of the element:

C: 1.1999 g / 12.011 g mol^{-1} = 0.0999 mol

H: 0.3021 g / 1.0079 g mol^{-1} = 0.2997 mol

The final step is to divide each of these two values by the smallest number, in this case this is the number of moles of carbon:

C: 0.0999 mol / 0.0999 mol = 1

H: 0.2997 mol / 0.0999 mol = 3

We therefore have a ratio of 1 carbon atom to 3 hydrogen atoms, thus the empirical formula for this hydrocarbon is CH_{3}.

The first step here is to determine the mass of C in CO_{2} and the mass of H in H_{2}O. This is done by dividing the atomic mass by the molecular mass and then multiplying by the mass of compound produced.

For C:

(12.011 g / 44.009 g) x 4.40 g = 1.1999 g

For H:

(1.0079 x 2 / 18.0148 g) x 2.70 g = 0.3021 g

The next step is to convert these masses into moles. This is done by dividing the mass by the relative atomic mass of the element:

C: 1.1999 g / 12.011 g mol^{-1} = 0.0999 mol

H: 0.3021 g / 1.0079 g mol^{-1} = 0.2997 mol

The final step is to divide each of these two values by the smallest number, in this case this is the number of moles of carbon:

C: 0.0999 mol / 0.0999 mol = 1

H: 0.2997 mol / 0.0999 mol = 3

We therefore have a ratio of 1 carbon atom to 3 hydrogen atoms, thus the empirical formula for this hydrocarbon is CH_{3}.

There are two ways to work this out, one is a lot simpler than the other, I will post the simpler method here.

The relative formula mass of magnesium oxide, MgO (40), is just a sum of the mass of magnesium (24) and the mass of oxygen (16).

24 + 16 = 40

We are told that 24 g of Mg produces 40 g of MgO when it reacts with 16 g of oxygen. We only need 10 g of MgO, 1 quarter of that given in the question.

In order to produce 10 g of MgO, we therefore only need a quarter of what we needed to make 40 g of MgO.

So, we would only need 24 g / 4 = 6 g of Mg to produce 10 g of MgO.

There are two ways to work this out, one is a lot simpler than the other, I will post the simpler method here.

The relative formula mass of magnesium oxide, MgO (40), is just a sum of the mass of magnesium (24) and the mass of oxygen (16).

24 + 16 = 40

We are told that 24 g of Mg produces 40 g of MgO when it reacts with 16 g of oxygen. We only need 10 g of MgO, 1 quarter of that given in the question.

In order to produce 10 g of MgO, we therefore only need a quarter of what we needed to make 40 g of MgO.

So, we would only need 24 g / 4 = 6 g of Mg to produce 10 g of MgO.