Prelim Chem Thread (1 Viewer)

[Snowflek]

Ohh i see thank you

 

[Snowflek]

what mass of silver chloride is obtained by adding excess silver nitrate solution to a 10.0mL volume of 0.10 mol L CacL2 solution?

 

[Suu]

To start off, find how many mols of Calcium Chloride there is in the solution.
Multiply the Molarity by how much solution there actually is. (1/10 *10/1000). There are 0.001 mols of CaCl2 in this solution.
(and also, 0.001 moles of Cl2).
Cl has a charge of -1. whilst Ag has a charge of +2.
We know that Ag and Cl will combine in the ratio 1:2 respectively. Therefore. 0.001 mole of AgCl2 will be formed.
AgCl2 weighs 107.8682+2*35.45 grams per mol. (178.7682g per mol).
178.7682 * 0.001 =0.1787682g, and the resulting mass of Silver Chloride will weigh this amount.

(I prematurely apologize for any mistakes that may, or may not be present in this answer. I have the cold at the moment.)

 

[Snowflek]

Thank you for helping me even though you have a cold, really hope you get better
Im kinda confused at the end though.. So ill go through the steps i did.(This is a multiple choice and the way did it and i did have both different answers but both answers are on the list.
My way: I wrote the equation out CaCl2 +2AgNO3 ---> 2AgCl + Ca(NO3)2
So the CaCl2 : 2AgCl. Ratio= 1:2
using c=n/v to calculate the mole of CaCl2
- 0.10 * 0.01 =0.001. Then i times 0.001 by 2 since the ratio is 1:2 and 0.001 is the mole of CaCl2
- 0.002 *(107.87 + 35.5) = 0.28674 (answer being B)
If i didnt times it by 2 it would still be 0.001 so
- 0.001 *(107.87 +45.5) =0.14337 (answer being A)

One of these are the answers
a) 0.143g
b)0.287G
c)14.3 g
d)143

 

[ml125]

Thank you for helping me even though you have a cold, really hope you get better
Im kinda confused at the end though.. So ill go through the steps i did.(This is a multiple choice and the way did it and i did have both different answers but both answers are on the list.
My way: I wrote the equation out CaCl2 +2AgNO3 ---> 2AgCl + Ca(NO3)2
So the CaCl2 : 2AgCl. Ratio= 1:2
using c=n/v to calculate the mole of CaCl2
- 0.10 * 0.01 =0.001. Then i times 0.001 by 2 since the ratio is 1:2 and 0.001 is the mole of CaCl2
- 0.002 *(107.87 + 35.5) = 0.28674 (answer being B)
If i didnt times it by 2 it would still be 0.001 so
- 0.001 *(107.87 +45.5) =0.14337 (answer being A)

One of these are the answers
a) 0.143g
b)0.287G
c)14.3 g
d)143

The answer is B - you need to multiply the moles of CaCl2 by 2 due to the 1:2 ratio.

 

[Snowflek]

Yeah that what i was thinking thanks!

 

[frog1944]

A compound is found to contain 23.3% magnesium, 30.7% sulfur and 46.0% oxygen. What is the empirical formula for this compound?

 

[jazz519]

1st step is to assume a mass normally 100 grams is the easiest as the mass of the components will just equal their percentage

So this means 23.3 grams of Mg, 30.7 grams of S, 46 grams of O

Convert these to moles, thus n(Mg) = 23.3/24.31 = 0.958453314 moles, n(Sulfur) = 30.7/32.07 = 0.9572809479 moles, n(oxygen) = 46/16 = 2.875 moles

Then you have to divide each mole value by the smallest moles you found so in this case Sulfur, so you get mg = 1, Sulfur = 1 and oxygen = 3 (when rounded to nearest whole number). Thus, the empirical formula is essentially these atom amounts in the compound you have found so it is MgSO3

 

[frog1944]

Awesome! Thank you so much. So for this question "What is the empirical formula for a compound containing 38.8% carbon, 16.2% hydrogen and 45.1% nitrogen." Would it be CNH_4 ?

 

[laloola]

^I got 5 for Hydrogen...

 

[jazz519]

^I got 5 for Hydrogen...

Yeah same

 

[frog1944]

Yeah CNH_5 sorry, I didn't round it up. Thanks

 

[Snowflek]

Hello so i dont get why we should assume its 100g when all you did was just change % to G. Cause like i dont see the 100grams used later on in the question

 

[jazz519]

You can assume any mass it doesn't really matter it could be like 500 grams but u will always end up with the same number of atoms in the empirical formula cause it's a ratio and same answer. I chose 100 grams because it is the simplest to do, like if you have a percentage that is overall equal to 100% so using 100 grams means the percentage will be equal to the grams, just making the process in these problems easier.

 

[Snowflek]

If im not mistaken, the second question is like 100.1 grams. But like i still dont get the point of the grams. All you did was change % to g and then find the moles and divide. Like i dont see you using the 100grams later on

 

[pikachu975]

If im not mistaken, the second question is like 100.1 grams. But like i still dont get the point of the grams. All you did was change % to g and then find the moles and divide. Like i dont see you using the 100grams later on

Because the empirical formula will be the same no matter how many grams of something you have, so you just use 100 grams to make it easier to convert percentage

 

[Snowflek]

I dont get it, all he did was change the percentage to grams. He didnt do anything about the 100 grams? Sorry im very confused atm

 

[pikachu975]

I dont get it, all he did was change the percentage to grams. He didnt do anything about the 100 grams? Sorry im very confused atm

The 100 grams was just used to get the empirical formula, which is the ratio between the elements for ANY AMOUNT OF GRAMS. You can use 12389123821 grams of the chemical if you want and still get the same empirical formula. How else are you going to get the mass of each element without assuming an amount of grams (100 grams is the easiest for percentage reasons)?

 

[Snowflek]

23.3% magnesium, 30.7% sulfur and 46.0% oxygen
So like lets say 100grams again
He finds 23.3% of 100grams which is 23.3 grams of mg right?

 

[pikachu975]

23.3% magnesium, 30.7% sulfur and 46.0% oxygen
So like lets say 100grams again
He finds 23.3% of 100grams which is 23.3 grams of mg right?

Yeah