Question

I have this question... I'm trying to work on it first so you can see my work so it will be up soon.
Q: Calculate the % iron in an unknown sample if 0.3400 g unknown was weighed, and 31.45 mL of 0.01788 M K2Cr2O7 was used to titrate to the endpoint. Make sure you use the coefficients from the reaction balanced in problem 2 above.

Answer 1

\[6 Fe^{+2} + Cr-{2}O_{7}^{2-} + 14 H+\rightarrow 6 Fe^{+3} + 2 Cr^{+3} + 7 H_{2}O\]

Answer 2

my first thought was to do this first
\[\frac{.01788 M K_{2}Cr_{2}O_{7}}{1}*\frac{1}{.031450L}=.568521M\]

Answer 3

or should I be trying to get moles instead?

Answer 4

can you please rewrite the reaction formula? You dont really need to

How to solve:
1.
Use this info to find moles 31.45mL of 0.01788 M K2Cr2O7, using the formula:
Molarity = moles/Volume
Volumes has to be in liters so multiple 31.45mL by (1L/1000mL) to convert to liters

2.
Use your reaction formula,

to find moles of iron

3. convert moles to grams using

Moles = grams/molecular mass

4. Use grams of iron found in step 3 and 0.3400g (the weight of the unknown) in the following formula:

% iron in unknown = (Grams of Iron/Grams of unknown)*100

Answer 5

\[\frac{.01788M}{1L}*.031450L=5.626260*10^{-4}moles\]

Answer 6

Can you take a screen shot of the formula or rewrite it because what you provided is pretty illegible

Answer 7

really how so it looks perfectly fine for me.

Answer 8

\[6 Fe^{+2} + Cr_{2}O_{7}^{2-} + 14 H+\rightarrow 6 Fe^{+3} + 2 Cr^{+3} + 7 H_{2}O\]
better?

Answer 9

It looks like a mess to me are you sure that is exactly how it is written in your homework. I provided the steps to solve your problem above.

Answer 10

Is that exactly how it is written?

Answer 11

I'm confused because there is no K2Cr2O7 in the equation. yes that is exactly how it is written.

Answer 12

You just need to use gravimetric factor to find moles of Cr2O^2 in K2Cr2O7

Answer 13

I will edit step 1

1.
a) Use this info to find moles 31.45mL of 0.01788 M K2Cr2O7, using the formula: Molarity = moles/Volume

Volumes has to be in liters so multiple 31.45mL by (1L/1000mL) to convert to liters

b) use gravimetric factor with the following equation using moles of K2Cr2O7 found above.

\[(Moles of K2Cr2O7)*\frac{molecular\; mass\; of\; Cr_2O_7^{-2}}{Molecular\; mass\; of\; K_2Cr_2O_7} = moles\; of\; Cr_2O_7^{-2}\]

Answer 14

In step 2. All you have to do is divide moles of Cr2O7^-2 by 1 and then multiply them by 6 to get moles of Fe^2+ then convert to grams of Iron using:

moles = grams/molecular mass

Answer 15

If you dont know how to find molecular mass look up how to find molecular mass in youtube

Answer 16

That is all the information you need to solve this problem

Answer 17

\[5.626260*10^{-4}moles of K_2Cr_2O_7^{2-}*\frac{215.985000}{294.181000}=4.130749*10^{-4}molesCr_{2}O_{7}^{2-}\]

Answer 18

\[4.130749*10^{-4}molesCr_{2}O_{7}^{2-}*\frac{6molesFe^{2+}}{1molesCr_{2}O_{7}^{2-}}\]
\[=.002478molesFe^{2+}\]
\[.002478molesFe^{2+}*\frac{55.845gFe^{2+}}{1molesFe^{2+}}\]
\[.138384g Fe^{2+}\]

Answer 19

\[\frac{.3400}{.138384}*100=\]
this isn't right @Australopithecus
i didn't work

Answer 20

0.01788*(31.45/1000) = 0.000562326mol
http://www.wolframalpha.com/input/?i=0.01788*%2831.45%2F1000%29

0.000562326mole*(215.988/294.185) = 0.00041285mol
http://www.wolframalpha.com/input/?i=+0.000562326*%28215.988%2F294.185%29

0.00041285mol*6 = 0.0024771mol
http://www.wolframalpha.com/input/?i=0.00041285mol*6

0.0024771mol*55.845g/mol = 0.13833g
http://www.wolframalpha.com/input/?i=0.0024771mol*55.845g%2Fmol

(0.13833/0.138384)*100 = 99.96%
http://www.wolframalpha.com/input/?i=+%280.13833%2F0.138384%29*100

Answer 21

Molecular mass of Cr2O7 is
http://www.wolframalpha.com/input/?i=Cr2O7+molecular+mass

Molecular mass of K2Cr2O7 is
http://www.wolframalpha.com/input/?i=K2Cr2O7+molecular+mass
294.185 daltons = 294.185g/mol

Answer 22

My method is correct I think you just made a mistake in the calculation

Answer 23

im with you up till here:
0.13833/0.138384)*100 = 99.96% where did you get the .138384 from?

Answer 24

@Australopithecus

Answer 25

should it be .3400?

Answer 26

I got 41% that way

Answer 27

oh yeah you are right

Answer 28

0.3400 g is the mass of the unknown my mistake

Answer 29

(0.13833g/0.3400g)*100 = 40.68%
http://www.wolframalpha.com/input/?i=%280.13833%2F0.3400%29*100+


You are right I was looking at your solution for some reason even though I knew it was wrong. :S

Answer 30

that is where I got 0.138384

Answer 31

Also I should have noticed you had the ratio flipped the wrong way

Answer 32

Dont get confused by my mistakes just follow my steps, if you do it exactly as i have written out you wont have any problems, but pretty sure the answer is 40%