How to calculate increase in mass of a metal when heated in oxygen?

[Mxrkko]

If a 1.00g sample of lithium was heated in oxygen until it was completely converted to an oxide, what increase in mass would you expect for the metal? Identify two factors that determine the percentage increase.

[jakesilove]

Hey hey!

I've attached my answer below; I hope it helps! If you have any further questions, feel free to post it here. My general recommendations for any maths questions is the ALWAYS START BY WRITING THE CHEMICAL EQUATION!



Jake

[Mxrkko]

Thank you so much! It makes complete sense now. I forgot that you can actually rearrange the moles = mass / molar mass formula to find the mass itself. Just to double check that i'm going along the right path, here's my working out for the question if it were to be applied to 1.00g of beryllium instead:

Balanced chemical equation: 2Be + O2 = 2BeO

Moles = 1.00/9.01 = 0.1109877913

Since it is the ratio of 2:2, beryllium oxide would also be 0.1109877913 moles

Now the moles = mass/molar mass formula is rearranged to mass= moles x molar mass

Mass of beryllium oxide =0.1109877913 x 25.01
                                       = 2.78g

2.78g - 1.00g = 1.78g

Therefore, an increase of 1.78g

[jakesilove]

Thank you so much! It makes complete sense now. I forgot that you can actually rearrange the moles = mass / molar mass formula to find the mass itself. Just to double check that i'm going along the right path, here's my working out for the question if it were to be applied to 1.00g of beryllium instead:

Balanced chemical equation: 2Be + O2 = 2BeO

Moles = 1.00/9.01 = 0.1109877913

Since it is the ratio of 2:2, beryllium oxide would also be 0.1109877913 moles

Now the moles = mass/molar mass formula is rearranged to mass= moles x molar mass

Mass of beryllium oxide =0.1109877913 x 25.01
                                       = 2.78g

2.78g - 1.00g = 1.78g

Therefore, an increase of 1.78g

That's absolutely perfect! Make sure to put states on your chemical equations (which I'm sure you would have, it you could write it out properly), but otherwise the working out is fantastic. I'm also glad that you restricted your answer to three significant figures: whilst they only sometimes take off marks for getting sig figs wrong, doing it right every time means you'll never lose marks.

Keep slugging through the content, and feel free to ask me questions any time!

Jake

[Mxrkko]

Will do! Thank you again for the help

[Mxrkko]

Hey Jake! I was wondering if you could help me break down the second part of this question:

Describe two modern day situations where chemical reactivity is an important factor in deciding which metal to use. Assess the importance of reactivity relative to other factors in making the decision.

I don't really understand what the second part of the question is asking?

[jakesilove]

Hey Jake! I was wondering if you could help me break down the second part of this question:

Describe two modern day situations where chemical reactivity is an important factor in deciding which metal to use. Assess the importance of reactivity relative to other factors in making the decision.

I don't really understand what the second part of the question is asking?

Hey!

The second half of that question is certainly a weird one. I think the sort of thing that its asking you to discuss is cost, weight, maybe conductivity etc.

I don't know any specifics, and I don't think you need to either. I would be talking about the importance of ensuring metals are not reactive, if needing to be used for long periods of time. However, weighing up that factor with things like cost (if the metal is super expensive, maybe it is better that it corrodes slightly more quickly), weight (depending on the purpose of the instrument, this could play a factor) etc. etc.

You should only need to answer very generally, identifying important factors, stating that they're important, but not drawing any specific conclusions. This is one of the strangest Chemistry questions I've ever seen.

Let me know if you need anything else! I know my answer was very vague, but that's really just because you're answer to the question will also be super vague.

Jake