[quote=Haishiro]1. 250ppb = 250*10^-9 250*10^-9 * 1.0 L = 250 * 10^-9 L KI Find the density of KI (g/L) to get g of KI
2. You don't need to use sq. rt. in addition or scalar multiplication of uncertainties. You can think of it as: .1 + .1 + .1 + .1 + .1 + .1 + .1 + .1 + .1 + .1 = +/-1.0mL as the absolute uncertainty
3. FeCl3 + KCl + AgNO3 -> AgCl + KNO3 + FeCl3 5.6878 M AgNO3 = 5.6878 mol / L AgNO3 5.6878 mol/L * 126.22 mL* (1L/1000mL) = 0.71791412 mol AgNO3 -> have 0.71791412 mol KCl -> have 0.71791412 mol K From there, find how many g K is equal to 0.71..mol K using the molar mass of K, then express K in ppm.[/quote]
Thanks, I figured you needed to know the density of the aqueous KI solution to find #1. I can't see any other way you could go by doing it.
I was having trouble with #3 in figuring out the chemical equation from the reactants. After that the stoichiometric calculations are simple. Thanks.
I just feel strange about simply adding the uncertainties in #2 since my book says to use the error of propagation for a very similar example. I'll read into it further I guess.
[quote=TheDarkHero3]I am pretty sure its because you want it in a percent form, you multiply it by 100, and then you multiply it by an additional 10 due to the fact that we are measuring 1000ml/100ml=10 times.[/quote]
I can understand why you'd want to multiply it by 10 (the measurement needs to be taken 10 times) but since it's random error, shouldn't you need to take the square root since you have to calculate propagation of error? That's what's messing me up.
I'm doing
total error, e = sq.rt(10 * (0.1/100)^2) and it's not coming out right.
[quote=SunsetDews]If you don't mind me asking, how did you get that?
I know (0.10 / 100.0) is the relative uncertainty for one measurement, but why'd you multiply it by 1000?[/quote] I am pretty sure its because you want it in a percent form, you multiply it by 100, and then you multiply it by an additional 10 due to the fact that we are measuring 1000ml/100ml=10 times.
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[quote=Haishiro]1. 250ppb = 250*10^-9
250*10^-9 * 1.0 L = 250 * 10^-9 L KI
Find the density of KI (g/L) to get g of KI
2. You don't need to use sq. rt. in addition or scalar multiplication of uncertainties.
You can think of it as:
.1 + .1 + .1 + .1 + .1 + .1 + .1 + .1 + .1 + .1 = +/-1.0mL as the absolute uncertainty
or (relative uncertainty):
0.1mL/100mL * 10 = 0.01 = 1% relative uncertainty
3.
FeCl3 + KCl + AgNO3 -> AgCl + KNO3 + FeCl3
5.6878 M AgNO3 = 5.6878 mol / L AgNO3
5.6878 mol/L * 126.22 mL* (1L/1000mL) = 0.71791412 mol AgNO3
-> have 0.71791412 mol KCl -> have 0.71791412 mol K
From there, find how many g K is equal to 0.71..mol K using the molar mass of K, then express K in ppm.[/quote]
Thanks, I figured you needed to know the density of the aqueous KI solution to find #1. I can't see any other way you could go by doing it.
I was having trouble with #3 in figuring out the chemical equation from the reactants. After that the stoichiometric calculations are simple. Thanks.
I just feel strange about simply adding the uncertainties in #2 since my book says to use the error of propagation for a very similar example. I'll read into it further I guess.
@above:
How did you get 1 kg = 1L?
250.0 ppb =
250g/1,000,000,000g = 250mg/1,000,000g = 0.250mg/1,000g = 0.250mg/1L (1,000g = 1Kg = 1L)
So, we need 0.250mg of Iodide.
Find how much grams of KI we need to have 0.250mg of I.
I = 126.9g/mol
KI = 166g/mol
166g/126.9g = 1.31
So, you have 1.31g of KI for every gram of I.
0.250mg * 1.31 = 0.328mg of KI
[quote=TheDarkHero3]I am pretty sure its because you want it in a percent form, you multiply it by 100, and then you multiply it by an additional 10 due to the fact that we are measuring 1000ml/100ml=10 times.[/quote]
I can understand why you'd want to multiply it by 10 (the measurement needs to be taken 10 times) but since it's random error, shouldn't you need to take the square root since you have to calculate propagation of error? That's what's messing me up.
I'm doing
total error, e = sq.rt(10 * (0.1/100)^2) and it's not coming out right.
250.0 ppb = 250.0/1,000,000,000
1L = 1000 mL
1000 mL/100 mL = 10 mL
10 mL x 10 mL = 100 mL
1000 mL +/- 100
That's all I got
[quote=SunsetDews]If you don't mind me asking, how did you get that?
I know (0.10 / 100.0) is the relative uncertainty for one measurement, but why'd you multiply it by 1000?[/quote]
I am pretty sure its because you want it in a percent form, you multiply it by 100, and then you multiply it by an additional 10 due to the fact that we are measuring 1000ml/100ml=10 times.
[quote=DarkEternity]2. (0.10 / 100.0) * 1000[/quote]
If you don't mind me asking, how did you get that?
I know (0.10 / 100.0) is the relative uncertainty for one measurement, but why'd you multiply it by 1000?
2. (0.10 / 100.0) * 1000