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HL Chemistry- Paper 1 (questions) help

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Hi guys,

 

So I started HL Chemistry about 2 months ago, and so far I think it's okay, except for Paper 1. We just finished Atomic Structure (Unit 2) and now are doing Stoichiometry (Unit 1), and I always find Paper 1 harder than 2. The reason being, in Paper 2 you're allowed calculators, but in Paper 1 you're not. Given that, our teacher told us that Paper 1 will have 'easy' numbers thus making mental calculations easy. However, from the tests/quizzes I've had for Paper 1 (especially for Unit 1: Stoichiometry) seem insanely difficult. I mean literally we're asked to convert mass to moles, moles to mass, moles to molecules, molecules to moles, mass to molecules, molecules to mass, etc. without calculators! How is that even possible? By all means, this is not 'easy' (rather VERY hard) as during these conversions decimals and exponents are being dealt with (in terms of addition/multiplication/division). The mental math skills required for Paper 1, from what I've experienced (Paper 1 tests/quizzes), seem far, FAR above the average human being.

 

I know I sound lame, but this is what I've experienced. Any help/advice will be greatly appreciated.

 

Thank you.

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Most of the problems that ask you to calculate something in Paper 1 will be very very simple calculations that reduce significantly or only have one or two steps to them. You don't have to perform mental steps, you can write down your thought process on paper to help you work through the problem if you need to. Also, in Paper 1 generally it's okay to round atomic masses to integers: a hydrogen atom can be thought of having a mass as 1 amu instead of 1.008, carbon can be rounded to 12 instead of 12.01, and so forth.

 

In general, you should be comfortable with basic algebra and know the basic rules for exponents and division, it will help a lot. If you can't solve a problem, you can usually eliminate some of the choices based on the answer itself if you know that it can't be a possible answer, and make an educated guess. If you work out the math and you get an answer that isn't exactly like one of the choices, but is relatively close, you should go with whichever is closest to your answer (or rework the math if you have time).

 

But seriously, you don't have to do everything mentally, write it out in the white space on the exam.

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I don't see how the calculations are simple. I do understand the entire rounding idea, but sometimes when converting mass to moles, moles to mass, moles to molecules, molecules to moles, mass to molecules, and molecules to mass, we are multiply or dividing numbers by decimals and exponents (for example, Avagadro's number- 6.022x10^23). Even if I do it on a paper, it can be pretty crazy.

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I don't see how the calculations are simple. I do understand the entire rounding idea, but sometimes when converting mass to moles, moles to mass, moles to molecules, molecules to moles, mass to molecules, and molecules to mass, we are multiply or dividing numbers by decimals and exponents (for example, Avagadro's number- 6.022x10^23). Even if I do it on a paper, it can be pretty crazy.

 

You should give us a concrete example on a question that you are struggled with. Otherwise, we can only give you general advice.

The tricks are basically to round up, to not use decimals, to use exponents, and to make approximate estimations. These estimations will help you eliminate one or more of the choices in a multiple-choice question.

 

For example: What is the molar mass of an unknown compound if 91.12 × 10^21 molecules of the same compound weigh 6.22 g?

 

Now the tip here is to first round up all the numbers appeared in the question (remember to always use scientific notations for very large or very small numbers!) :

- 91.12 × 10^21 gets rounded up to 9 × 10^22

- Avogadro's number becomes 6 × 10^23

- 6.22 becomes 6

 

Next, calculate the number of moles that 9 × 10^22 molecules represent by dividing this with Avogadro's number!

- Calculate the exponents first: 10^22 ÷ 10^23 = 10^(22 - 23) = 10^(-1) = 1/10

- Now, 9 ÷ 6 = 3/2

- Multiply them both, you get 3/2 × 1/10 = 3/20 moles (which is basically 'fraction multiplication')

 

The final step now is to calculate the molar mass:

- Using the result above that "3/20 moles represent 6g", it's clear to see that:

                                    3/20 mol --> 6g

                                      1    mol --> ?g

- Now using cross-multiplication, we get:

                                    ? = 6 × 1 ÷ 3/20 = 40g (which again is basically 'fraction multiplication')

 

Since we have done many round-ups in our calculation, the answer would not be exactly 40g (in fact the real answer is 39.7g). But now all you need to do is pick the choice that is closest to 40g!

 

This example i gave here might be a little bit easier than the ones in the real exam; but anyhow, if you break a question down into many steps (in which calculations for each step are easy), you will never get it wrong.

 

Note that as you repeat this process many times with many questions, your 'mentally-calculating' skills will get so much better and faster. So practice often with calculations of these broken-down steps until you really master them. By then, you'll probably think that calculators are so useless for these types of questions ;)

 

Hope that helps! Cheers :)

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Hi,

 

That helps a lot. Although I still don't feel quite confident with all these calculations in my head and on paper (even with rounding), but I guess practice makes perfect. Thanks for this example!

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