Chemistry math confuses me

[Mole]

I don't understand the math used in my chemistry book at all. Everything about significant figures seems illogical to me - like, for instance, multiplying 3.253 and 1.11, and only being able to get an answer to two decimal places because the less certain of the two numbers only has three significant figures.

Here's something right out of my Holt Chemistry book:

[...]copper atoms have an average mass of only 1.0552 x 10^-25 kg.
Each penny in Figure 23 [two copper pennies] has an average mass of 3.13 x 10^-3 kg and contains copper. How many copper atoms are there in one penny? Assuming that a penny is pure copper, you can find the number of copper atoms by dividing the mass of the penny by the average mass of a single copper atom or by using the following conversion factor:

1 atom Cu/(1.0552 x 10 ^ -25 kg)

(3.13 x 10^-3 kg) x ([1 atom Cu]/[1.0552 x 10^-25 kg]) = 2.97 x 10^22 Cu atoms

In the book 1 atom Cu is displayed over 1.0552 x 10^-25 kg, and the kg's are crossed out. I don't see how it's done at all. 3.13 divided by 1.0552 is about 2.97, but I don't see the above equation as 3.13/1.0552. I see (3.13 x 10^-25) x (1/1).

 

[Deleted member 4993]

I don't understand the math used in my chemistry book at all. Everything about significant figures seems illogical to me - like, for instance, multiplying 3.253 and 1.11, and only being able to get an answer to two decimal places because the less certain of the two numbers only has three significant figures.

Here's something right out of my Holt Chemistry book:

[...]copper atoms have an average mass of only 1.0552 x 10^-25 kg.
Each penny in Figure 23 [two copper pennies] has an average mass of 3.13 x 10^-3 kg and contains copper. How many copper atoms are there in one penny? Assuming that a penny is pure copper, you can find the number of copper atoms by dividing the mass of the penny by the average mass of a single copper atom or by using the following conversion factor:

1 atom Cu/(1.0552 x 10 ^ -25 kg)

(3.13 x 10^-3 kg) x ([1 atom Cu]/[1.0552 x 10^-25 kg]) = 2.97 x 10^22 Cu atoms

In the book 1 atom Cu is displayed over 1.0552 x 10^-25 kg, and the kg's are crossed out. I don't see how it's done at all. 3.13 divided by 1.0552 is about 2.97, but I don't see the above equation as 3.13/1.0552. I see (3.13 x 10^-25) x (1/1).

Suppose you have box with steel balls and the whole box weighs 30 kg.

Suppose you know each ball weighs 6 Kg - and the empty box weighs nothing (say negligible for this problem).

Then how many balls are there in the box?

 

[mmm4444bot]

Hi Mole:

You wrote,

... I don't see the above equation as 3.13/1.0552 ...

You also wrote,

... I see (3.13 x 10^-25) x (1/1).

Hmmm. Why do you see this product?

When I look at the left side of that equation, I see (3.13 x 10[sup:3rbld1wd]-3[/sup:3rbld1wd]) x (1/1)

If your post is an attempt to tell us that you do not understand the concept of conversion factors, then that is something else entirely.

Conversion factors belong to a process known as "dimensional analysis", whereby you cancel UNITS instead of NUMBERS.

Multiplying by a ratio that equals (1/1) allows us to do this without changing the NUMERICAL value.

The topic known as "significant figures" is something else again. CHEMISTS (i.e., books and teachers) are known for sometimes using their own tailored set of rules to determine significant figures. These rules do not always agree with pure mathematics. I would need to see what you've been taught.

Please let us know about what exactly you are trying to ask. Numerous partial statements make it hard for me to figure out.

~ Mark

PS: Here's an example of how to create a conversion factor to convert 46350 km/hr to mi/sec; at this point, I don't know if it helps you...

\(\displaystyle \frac{km}{hr}\;\times\;\frac{hr}{sec}\;\times\;\frac{mi}{km}\)

The UNITS hr and km cancel, resulting in mi/sec.

Therefore, our conversion factor comes from this part:

\(\displaystyle \frac{hr}{sec}\;\times\;\frac{mi}{km}\)

Now we stick in the numbers to go with those units, making SURE that each ratio equals ONE.

\(\displaystyle \frac{1 hr}{3600 sec}\;\times\;\frac{0.621371 mi}{1 km}\)

The conversion factor is 0.0001726, or 1.726EE-4 in Scientific Notation (where EE stands for "times 10 raised to the power of").

\(\displaystyle \frac{46350 km}{hr}\;\times\;0.0001726\frac{mi|hr}{sec|km}\;=\;8 \frac{mi}{sec}\)

My edits: deleted stuff about ratios; corrected ambiguous language; keep finding misspeled words, too ...

 

[Denis]

1 atom Cu/(1.0552 x 10 ^ -25 kg)
(3.13 x 10^-3 kg) x ([1 atom Cu]/[1.0552 x 10^-25 kg]) = 2.97 x 10^22 Cu atoms
In the book 1 atom Cu is displayed over 1.0552 x 10^-25 kg, and the kg's are crossed out. I don't see how it's done at all. 3.13 divided by 1.0552 is about 2.97, but I don't see the above equation as 3.13/1.0552. I see (3.13 x 10^-25) x (1/1).

NOTE: please use * (not x) as multiplication sign; now standard

1 atom Cu/(1.0552 x 10 ^ -25 kg) simply means 1 /(1.0552 * 10 ^ -25) ; like, dividing by k is same as multiplying by 1/k
(I have no idea why your book does that: simply silly to me)

So: (3.13 * 10^-3) * [1 /(1.0552 * 10 ^ -25)]
= (3.13 * 10^-3) / (1.0552 * 10 ^ -25)
= (3.13 * 10^22) / 1.0552
= 2.96626.... * 10^22

That's (rounded) 29 662 623 199 393 479 909 291 :shock:

Ya'll ok now :?:

 

[mmm4444bot]

... You also wrote,

... I see (3.13 x 10^-25) x (1/1).

Hmmm. Why do you see this product?

When I look at the left side of that equation, I see (3.13 x 10[sup:ncxqiiaf]-3[/sup:ncxqiiaf]) x (1/1)

Oh, okay. After reading Denis' post, I can see where maybe the original poster made a typographical error.

Hey, Mole. Were you trying to tell us that you see the following?

(3.13 x 10[sup:ncxqiiaf]22[/sup:ncxqiiaf]) x (1/1)

Anyway, I'm still not sure what the original poster is trying to ask ...

~ Mark :?:

 

[Mole]

... You also wrote,

... I see (3.13 x 10^-25) x (1/1).

Hmmm. Why do you see this product?

When I look at the left side of that equation, I see (3.13 x 10[sup:3b8b33li]-3[/sup:3b8b33li]) x (1/1)

Oh, okay. After reading Denis' post, I can see where maybe the original poster made a typographical error.

Hey, Mole. Were you trying to tell us that you see the following?

(3.13 x 10[sup:3b8b33li]22[/sup:3b8b33li]) x (1/1)

Anyway, I'm still not sure what the original poster is trying to ask ...

~ Mark :?:

Yes. 1 atom Cu/(1.0552 x 10^-25) = 1, since 1 atom Cu=1.0552 x 10^-25. Am I missing the point of conversion factors?

 

[mmm4444bot]

... Am I missing the point of conversion factors?

You are the only person who really can answer this question.

I would like to know if you understand my example conversion factor for converting km/hr into mi/sec.

Do you see how the conversion factor contains all those units?

The "unit" on that conversion factor is 'mile-hours per kilometer-seconds".

This "unit" is not supposed to make sense to us if we examine it outside of the context of a conversion factor.

This "unit" is in place soley to get the desired units in the RESULT after we multiply.

If you cannot see how the units cancel during the use of this conversion factor to change km/hr into mi/sec, then, as Denis is fond of saying, "Only your teacher can help you."

Of course, we welcome additional (hopefully specific) questions if you would still like to sort out what's happening with conversion factors.

The conversion factor in your chemistry book is set up to change kg into number of atoms.

~ Mark

My edit: in red

 

[Deleted member 4993]

[/size]

Yes. 1 atom Cu/(1.0552 x 10^-25) = 1, since 1 atom Cu=1.0552 x 10^-25. Incorrect Am I missing the point of conversion factors?[/quote]

The correct statement would be

Mass of 1 atom of CU = 1.0552 x 10^-25 kg .................Don't forget units

So

Number of atoms in 1 kg of Cu = 1/(1.0552 x 10^-25) = 9.4769 * 10^(24) [notice no units - since we are talking about numbers]

Some times when you use this number - like in this problem - you need to attach a unit like atoms/kg to balance out other units.