# Question #81af3

##### 2 Answers
Mar 27, 2015

I looked at your lab and results and assuming you did the lab correctly there is probably a math error or a significant problem in your procedure (large % error)

Unfortunately, I have no idea what you actually did so it is very difficult to figure out where you went wrong. Some of my questions would be
1. How did you fill the flask with carbon dioxide?
2. Did you fill the flask twice or just measure the mass twice? Either way - how did you choose which value to use?
3. Where does the data come from? you say you attached calculations for the moles of CO2 section but I can't see them.

Would like to help but I need to know more - describe the procedure

Mar 27, 2015

You made some calculation errors, but you got an excellent result.

Here are the calculations for Experiment 1.

Part A

${m}_{\text{water" = m_"flask+water" – m_"flask+air" = "238.40 g – 99.1192 g" = "139.28 }} g$

${V}_{\text{flask" = V_"water" = (139.28 cancel("g water")) × "1 mL"/(0.99799cancel("g water")) = "139.56 mL}}$

${V}_{\text{CO₂" = V_"flask" = "139.56 mL}}$

${P}_{\text{CO₂" = (766.9 cancel("mmHg")) × "1 atm"/(760 cancel("mmHg")) = "1.0091 atm}}$

${T}_{\text{CO₂" = "(22.1+ 273.15) K" = "295.25 K}}$

${n}_{\text{CO₂" = (PV)/(RT) = (1.0091 cancel("atm") × 0.13956 cancel("L"))/(0.08206 cancel("L·atm·K⁻¹")"mol"⁻¹ × 295.25 cancel("K")) = "0.005 813 mol}}$

Part B

${V}_{\text{air" = V_"flask" = "139.56 mL}}$

${m}_{\text{air" = 139.56 cancel("mL air") × "0.001204 g air"/(1cancel("mL air")) = "0.1680 g air}}$

${m}_{\text{flask" = m_"flask+air" – m_"air" = "99.1192 g – 0.1680 g" = "98.9512 g}}$

${m}_{\text{CO₂"= m_"flask+CO₂" - m_"flask "= "99.2048 g – 98.9512 g" = "0.2536 g}}$

Part C

$M {M}_{\text{CO₂" = "0.2356 g"/"0.005 813 mol" = "43.63 g/mol}}$

Now go and repeat the calculations for Experiment 2.