Question

In 2006, professor galen supped, from the university of missouri columbia, was awarded a presidential green challenge award for his system of converting glycerin, C3H5(OH)3 a by-product of biodiesel production, to propylene glycol, C3H6(OH)2. Propylene glycol produced in this way will be cheap enough to replace the more toxic ethylene glycol that is the primary ingredient in automobile antifreeze (A) if 50.0 mL of propylene glycol has a mass of 51.80 grams what is the density? (B) To obtain the same antifreeze protection requires 76g of propylene glycol to replace each 62g of ethylene glycol.

Answer 1

Continuation in letter B: calculate the mass of propylene glycol required to replace 1.00 gal of ethylene glycol. The density of ethylene is 1.12 g/mL

(C) calculate the volume of propylene glycol, in gallons needed to produce the same antifreeze protection as 1.00 gal of ethylene glycol.

Answer 2

@FibonacciChick666 @watermelon14 Can i ask your help? :3 i've been solving this problem. But i can't trust my answer.

Answer 3

I am sorry I havnt learned that yet :(

Answer 4

Ohh. Okaay, thanks

Answer 5

@FibonacciChick666

Answer 6

@robtobey

Answer 7

For part a, you'll want to used the density formula again. The variable for density is usually \(\rho\), so I'll use that. Mass is \(m\) and volume is \(V\). Then \(\rho=m\div V\). \(51.80\ [g]\) is your mass. \(50.0\ [mL]\) is your volume.

Answer 8

http://hyperphysics.phy-astr.gsu.edu/hbase/dens.html#dens

Answer 9

Yeah, i got the answer in letter A 1.02 g/mL. Right? :3

Answer 10

I got \(1.036\ [g/mL]\). That would round to \(1.04\ [g/mL]\), to keep with your significant figures. Would you agree?

Answer 11

For part b, I don't see the question.

Answer 12

Nevermind! :P

Answer 13

No, if you divide 51.08 g and 50.0 mL. The answer would be 1.0216. Right? so if you round it off. It would be 1.02, right? Right? XD

Answer 14

Ohh, I'm sorry. You were right. XD i used 51.08 instead of 51.80. I'm sorry.

Answer 15

I figured it was calculator error for one of us! I'm making a table of the variables - to get organized. I'll post it in a moment!

Answer 16

\[\begin{matrix}\text{quantity}&\text{eth. gly.}&\text{prop. gly.}\\\rho\ [g/mL]&1.12&1.04\\V\ [gal]&1.00&?\end{matrix}\]And \(76\ [g]\text{prop.}\equiv62\ [g]\text{eth.}\) in terms of antifreeze protection..

Answer 17

what letter are you answering? :)

Answer 18

B.

And you don't know either masses. You don't know how much mass 1 gallon of ethylene glycol has. And you don't know how much mass of propylene glycol you'll need to have the same antifreeze effect.
\[\rho=m\div V\implies m=\rho\ V\]1 gallon is 3,785.41 milliliter, according to Google. So we'll use that as our volume.\[m=1.12\ [g/mL]\times 3,785.41\ [mL]=4,239.6592\ [g]\approx 4.24\times 10^4\ [g]\]

Answer 19

You already have A.

Answer 20

\[\begin{matrix}\text{quantity}&\text{eth. gly.}&\text{prop. gly.}\\\rho\ [g/mL]&1.12&1.04\\V\ [gal]&1.00&?\\m\ [g]&4.24\times 10^4&?\end{matrix}\]So, we know everything about ethylene glycol. And we can find out how many grams of propylene glycol we'll need to do the same antifreeze work.\[76\ [g\ prop]=62\ [g\ eth.]\]We have the grams of ethylene glycol, and we want the grams of propylene glycol.
Any idea? It's much like a conversion.

Answer 21

what letter is that?

Answer 22

\(m\) for mass.

Answer 23

you mean the grams of stuff? Or which letter of the problem?

Answer 24

which letter of the problem. :)

Answer 25

Oh, B! Let me know when you're caught up!

Answer 26

ok. :) i will.

Answer 27

A takes a little work, B takes a bit more work, and then C will be about as bad as A. :)

Answer 28

Haha.

Answer 29

:D

Answer 30

4.08x10^4g. I'm not pretty sure :/ @theEric

Answer 31

that's the volume of propylene glycol.

Answer 32

Let me check!

Answer 33

okay :)

Answer 34

Grams is a unit of mass. Did you mean that's the mass of propylene glycol?

Answer 35

that's gal. not g, i was wrong. Sorry

Answer 36

I didn't get there yet. Part B asks for mass. What did you get for that?

Answer 37

ohh wait. I will calculate it.

Answer 38

69.4 mL that's my answer for the mass of propylene glycol.

Answer 39

@theEric

Answer 40

\[76\ [g\ prop]=62\ [g\ eth.]\]\[m_{eth}=4.24\times 10^4\ [g eth]\]
So convert the grams of ethylene glycol to grams of propylene glycol - just like they're units of antifreeze ability! Do you know what I mean?

Answer 41

Hmmm?

Answer 42

Okay. I surrender, i can't get it. T_T

Answer 43

I can't get the answer.

Answer 44

Haha, all you had to say is that you didn't know what I mean. And I'll show you.

Answer 45

Haha. Stop reading my mind. XD yeaah, i admit it. :P

Answer 46

\[4.24\times 10^3\ [g\ eth]\times \frac{76\ [g\ prop]}{62\ [g\ eth]}\\=5.1974193548387096774193548387097\times 10^3\ [g\ prop]\\\approx 5.20\times10^3\ [g\ prop]\]

Answer 47

That would be B.

Answer 48

Ahhh! I get it :)

Answer 49

Next\[\begin{matrix}\text{quantity}&\text{eth. gly.}&\text{prop. gly.}\\\rho\ [g/mL]&1.12&1.04\\V\ [gal]&1.00&?\\m\ [g]&4.24\times 10^3&5.20\times 10^3\end{matrix}\]
1 unknown left. The volume of propylene glycol.
Use \(\rho=m\div V\implies V=m\div\rho\). You have \(m\) in [g] and \(\rho\) in g/mL, so you'll get your \(V\) in mL. But then you can convert, if you want.

Answer 50

Do you see what to do there?

Answer 51

yeah. :) wait. I'll try to answer it.

Answer 52

Okay! I'll be here!

Answer 53

5000 gal.

Answer 54

That doesn't seem right... Seems a little too high.

Answer 55

\(V=m\div \rho\)
\(m=5.2\times 10^3\ [g]\)
\(\rho=1.04\ [g/mL]\)
Plug it into a calculator and...

Answer 56

it's 5000 gal, the same with my answer.

Answer 57

i mean 5000 mL

Answer 58

wait, i'm wrong. I'll convert it to gallons.

Answer 59

I got it. @theEric 1.32 gal

Answer 60

@theEric are you still there? :3

Answer 61

Yep! I agree with 5000mL! I'm working on the conversion! My browser is being slow.

Answer 62

1.32 gallons seems reasonable, but I want to double-check the calculation for you.

Answer 63

I got the same as you, and that's C! :)

Answer 64

Yay! :D thank you so much eric. ^_^

Answer 65

You're very welcome. I hope it's all right! :)