A liquid solution of formaldehyde is 37% pure. This solution has a specific gravity of 1.037. The molecular weight of formaldehyde is 30.0. To 1.00 ml of this impure stock solution is added quantity sufficient total the total volume to 500 ml. If 1.00 ml of this solution is added to 4 ml of water in a final reaction, what is the molarity of formaldehyde in this final reaction tube?
So far: I have found the grams by doing
.001 L * 1.037 * .37 g/L = 3.84*10^-4
Then I found the M from here
3.84*10^-4 g * 30 moles/g = .0115 moles/ 1L = .0115 M
I don't thing this is correct because I didn't use any dilution or anything? Some direction on where I would use the dilutions would be helpful
I wouldn't go that route. I think this route is easier to see.
First, determine the molarity of the inital formaldehyde solution.
Thats 1.037 g/mL x 1000 mL x (37/100) x (1 mol/30 g) = about 12.8 M
Then use the dilution formula of
mL1 x M1 = mL2 x M2
For the first dilution you have
1 mL x 12.8 = 500 mL x M2
and M2 = 1*12.8/500 = approx 0.03 but you need a more accurate answer than that.
For the second dilution you have to assume that the volumes of additive; i.e., that 1 mL + 4 mL = 5 mL, then
1 mL x 0.03M = 5*M2 and solve for M2.
Another way to do the dilutions is
12.8 M x (1/500 ) = about 0.03
and 0.03 x 1/5 = ?
A third way is to look at the dilution factor.
You start with 12.8 M
You dilute 1 to 500 and again 1 to 5 so the total dilution is 500*5 = 2500 so the final concn is 1/2500 th the initial or 12.8/2500 = ?
Well, it seems like you're on the right track! Let's break it down a bit further.
First, let's calculate the number of moles of formaldehyde (CH2O) in the 1.00 ml of the impure stock solution:
Moles of CH2O = (0.001 L) x (0.37 g/L) / (30.0 g/mol) = 1.233 x 10^-5 mol
Now, since we're diluting this to a total volume of 500 ml, the number of moles remains the same. However, the final volume has changed, so we can calculate the concentration (molarity) in the final reaction tube as follows:
Molarity of CH2O = (1.233 x 10^-5 mol) / (0.500 L) = 2.466 x 10^-5 M
So, the molarity of formaldehyde in the final reaction tube is approximately 2.466 x 10^-5 M.
I hope this helps clarify things for you! And remember, if you need any further assistance, just let me know. I'm here to make chemistry a little more fun!
To find the molarity of formaldehyde in the final reaction tube, you need to take into account the dilution of the stock solution.
Let's break down the calculations step-by-step:
1. Calculate the number of moles of formaldehyde in the original 1.00 ml of the stock solution:
Moles of formaldehyde = volume (in L) × concentration (in M)
Moles of formaldehyde = 0.001 L × 0.37 M
Moles of formaldehyde = 3.7 × 10^-4 moles
2. Calculate the total number of moles of formaldehyde in the final 500 ml solution:
Total moles of formaldehyde = moles from stock solution × (final volume / initial volume)
Total moles of formaldehyde = (3.7 × 10^-4 moles) × (500 ml / 1 ml)
Total moles of formaldehyde = 0.185 moles
3. Calculate the molarity of formaldehyde in the final reaction tube:
Molarity of formaldehyde = moles of formaldehyde / volume (in L)
Molarity of formaldehyde = 0.185 moles / (1 ml + 4 ml) × (1 L / 1000 ml)
Molarity of formaldehyde = 0.037 M
Therefore, the molarity of formaldehyde in the final reaction tube is 0.037 M.
To find the molarity of formaldehyde in the final reaction tube, you will need to consider the dilution that occurs when adding the impure stock solution to water.
Let's break down the steps involved:
1. Calculate the number of moles of formaldehyde present in the impure stock solution:
- Given: Volume of stock solution = 1.00 ml
- Concentration of stock solution = 37% pure
- Density (specific gravity) of stock solution = 1.037
- Molecular weight of formaldehyde = 30.0 g/mol
Using the given information, you rightly calculated the mass of formaldehyde in the stock solution as 3.84 * 10^-4 g. Now, let's calculate the number of moles of formaldehyde in the stock solution:
- Moles = mass / molar mass
- Moles = 3.84 * 10^-4 g / 30.0 g/mol
- Moles = 1.28 * 10^-5 mol
2. Determine the dilution factor when adding the impure stock solution to make a final volume of 500 ml:
- Starting volume = 1.00 ml
- Final volume = 500 ml
The dilution factor is given by the ratio of the final volume to the starting volume:
- Dilution factor = Final volume / Starting volume
- Dilution factor = 500 ml / 1.00 ml
- Dilution factor = 500
3. Calculate the number of moles of formaldehyde in the final reaction tube:
- Moles in final reaction tube = Moles in stock solution * Dilution factor
- Moles in final reaction tube = 1.28 * 10^-5 mol * 500
- Moles in final reaction tube = 6.4 * 10^-3 mol
4. Finally, find the molarity of formaldehyde in the final reaction tube:
- Molarity = Moles / Volume
- Volume = 1.00 ml (from the given information)
- Molarity = 6.4 * 10^-3 mol / 1.00 ml
- Molarity = 6.4 * 10^-3 mol / 0.001 L (since 1 ml = 0.001 L)
- Molarity = 6.4 M
Therefore, the molarity of formaldehyde in the final reaction tube is 6.4 M.