I'm showing the procedure of the lab and what was asked. What I have question on is towards the end. Much help would be appreciated a lot!!
First I had to balance the equation given to me:
2 Na HCO3 (s) Δ over arrow (produces)
Na2 CO3 (s) + CO2 (g) + H2O (g).
From the lab observations:
A. Mass of evaporation dish & watch glass = 75.705 g
B. Mass of evaporating dish, watch glass, & NaHCO3 = 78.605
C. Mass of evaporating dish, watch glass, & Na2CO3 = 57.460 g
1.) Using your data, calculate the mass of Na2CO3 produced?
75.705g - 57.460 g = 18.245 g
2.) Calculate the mass of NaHCO3 that decomposed?
78.605 g - 75.705 g = 2.9
If correct sig figs, wouldn't the answer be 2.9000 or just 2.9?
Wouldn't this affect my answer if correct sig figs aren't correct?
4.) Using your data, calculate how many grams of Na2CO3, Theoretically should have been produced using stoichiometry? (Hint use your calculated amount of NaHCO3 to start the stoich... chart & refer to balanced.
baking soda Δ> carbon dioxide + water + sodium carbonate
2 NaHCO3 Δ CO2 + H2O + Na2CO3
Answer (I know this isn't right! Can someone break down the process including numbers of what to multiply with what etc.) :
2.900 g NaHCO3 x 1molNaHCO3/ 84.0059 gNaHCO3 x 1molNa2CO3/ 2 mol NaHCO3 x 105.988 g Na2CO3 / 1mol Na2CO3 = 1.829 g Na2CO3
Is that correct though?
Now #5 asks calculate percent yield & use experimental mass of Na2CO3 which is 18.245... & theoretical amount of Na2CO3 which is (1.829) BUT IDK IF THAT's RIGHT (please check my math!! THANKYOU SO MUCH...based on the stoichiometry...
18.245 / 1.829 x 100 = 997.5
That does not seem right though. so, where did I go wrong?
CHEMISTRY - bobpursley, Monday, February 22, 2016 at 9:24pm
2. It is good to five sig figures.
4, right, but obviously something else is wrong.
5. What is wrong is your mass of the final is wrong.
You can't start out with the empty apparatus mass greater the final apparatus weight with product.
CHEMISTRYHELPP!! BOBPURSLEY - Allie, Monday, February 22, 2016 at 9:37pm
I only did that ^ to get your attention. Thankyou for your help..
2. so is it 2.9000 because technically that's wrong cuz theres 3 digits on the right of the period right?
or is it 20.900 but that woudn't make sense. even if it was 02.900, the zero wouldn't matter. so what is the number may I ask?
so #4 is good?
#5 How do I fix it? Really confused at this point. Please help. Would really appreciate it.
Based on the information you provided, let's go through each question step by step:
1.) To calculate the mass of Na2CO3 produced, subtract the initial mass of Na2CO3 from the final mass of Na2CO3:
Final mass of Na2CO3 = Mass of evaporating dish, watch glass, & Na2CO3 - Mass of evaporating dish & watch glass
= 57.460 g - 75.705 g
= -18.245 g
Please note that the result is negative, which indicates an error in your measurements or calculations.
2.) To calculate the mass of NaHCO3 that decomposed, subtract the initial mass of NaHCO3 from the final mass of NaHCO3:
Mass of NaHCO3 decomposed = Mass of evaporating dish, watch glass, & NaHCO3 - Mass of evaporating dish & watch glass
= 78.605 g - 75.705 g
= 2.900 g
In terms of significant figures, the answer should be reported as 2.900 g, since there are three significant figures in the data given.
4.) To calculate the theoretical mass of Na2CO3 produced using stoichiometry, you need to use the balanced chemical equation and the molar masses of the compounds involved.
The molar masses are as follows:
NaHCO3: 84.0059 g/mol
Na2CO3: 105.988 g/mol
Using stoichiometry, the calculation would be as follows:
2.900 g NaHCO3 x (1 mol NaHCO3 / 84.0059 g NaHCO3) x (1 mol Na2CO3 / 2 mol NaHCO3) x (105.988 g Na2CO3 / 1 mol Na2CO3) = 1.830 g Na2CO3
So, the correct answer for the theoretical mass of Na2CO3 produced is 1.830 g, rounded to three significant figures.
5.) To calculate the percent yield, you need to divide the experimental mass of Na2CO3 by the theoretical mass of Na2CO3, and then multiply the result by 100.
Percent yield = (Experimental mass of Na2CO3 / Theoretical mass of Na2CO3) x 100
= (18.245 g / 1.830 g) x 100
= 997.81%
So, the correct answer for the percent yield is approximately 997.81%. Please note that your previous calculation did not have the correct theoretical mass of Na2CO3, which caused the incorrect result.
Based on the calculations you provided, let's go through each question one by one:
1) The calculation for the mass of Na2CO3 produced is correct. You subtracted the initial mass (57.460 g) from the final mass (75.705 g) to get 18.245 g.
2) For the mass of NaHCO3 that decomposed, you correctly subtracted the initial mass (75.705 g) from the final mass (78.605 g) to get 2.9 g. Considering significant figures, you are correct that it can be written as 2.9 g. However, if you are using the given masses with three significant figures, you should report the answer as 2.90 g.
4) To calculate the theoretical mass of Na2CO3 produced using stoichiometry, you need to use the balanced equation:
2 NaHCO3 Δ CO2 + H2O + Na2CO3
Given that you have 2.9 g of NaHCO3, you can use the following conversion factors:
2.9 g NaHCO3 x (1 mol NaHCO3 / 84.0059 g NaHCO3) x (1 mol Na2CO3 / 2 mol NaHCO3) x (105.988 g Na2CO3 / 1 mol Na2CO3)
This calculation gives you 1.829 g Na2CO3, which is the correct answer.
5) The percent yield calculation is typically done using the actual experimental yield and the theoretical yield. However, in this case, there seems to be an issue with the experimental mass of Na2CO3 you provided (18.245 g). As mentioned by Bobpursley, the final mass should not be greater than the initial mass.
To fix this, you should carefully review your lab observations and calculations to ensure you have the correct values for the masses. Once you have the correct experimental mass of Na2CO3, you can then calculate the percent yield using the formula:
Percent Yield = (Experimental Yield / Theoretical Yield) x 100
For example, if the correct experimental mass of Na2CO3 is 1.829 g (as calculated in question 4), the percent yield would be:
(1.829 g / 1.829 g) x 100 = 100%
Remember, the percent yield should always be less than 100% in most practical scenarios.