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!

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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.