Homework 95

8.41

Photosynthesis is a chemical reaction in plants used to produce oxygen and glucose. If a plant uses 6.3g CO₂, how much water (in grams) is needed to complete the reaction? How much glucose will be formed?

\[6CO_{2}+6H_{2}O+light\rightarrow C_{6}H_{12}O_{6}+6O_{2}\]

Solution/ part 1: To solve this you must first convert grams of CO₂ to moles of CO_2. From there you cross the mole bridge, which is the mole ratio between the molecules, this simply means that you will be going from moles of one compound to moles of a different compound. Finally, you must convert the moles of H₂0 to grams of H₂0. Then, you will plug all of those numbers into your calculator to get your solution.

Step 1- The first step to solving this problem is to convert the grams of CO₂ to moles of CO₂. To do this you will divide 1 mol CO₂ by the molar mass of CO₂. You find the molar mass by looking on the periodic table. You then multiply 6.3g CO₂ by the value you found which is \[\frac{1molCO_{2}}{44gCO_{2}}.\]

\[6.3gCO_{2}\frac{1molCO_{2}}{44gCO_{2}}\]

Step 2- Next, you need to cross the mole bridge. This is where you show the mole ratio between the molecules. This simply means that you will be going from moles of CO₂ to moles of H₂0. You do this because you ae trying to solve for how much water is needed to complete the reaction and you started with CO_2. The mole ratio of CO₂ to H₂0 is 6:6. If you look at the equation you will see that. You need to divide the moles H₂0 by moles of CO₂. You will take step 1 and multiply it with what you have found here which is \[\frac{6molH_{2}O}{6molCO_{2}}\]. This will give you

\[6.3gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}}*\frac{6molH_{2}O}{6molCO_{2}}\]

Step 3- Now you need to convert mol H₂0 to grams of H₂0. To do that you need to multiply by the molar mass of H₂0. Once again, you can find the molar mass on the periodic table. The molar mass of H₂0 is 18 grams per mole.\[\frac{18gH_{2}O}{1molH_{2}O}\]. Now that you have this you will this value and multiply it by what you found in step 2. This will give you

\[6.3gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}}*\frac{6molH_{2}O}{6molCO_{2}}*\frac{18gH_{2}O}{1molH_{2}O}\]

Step 4:

Now that you have all your values you just plug them into your calculator and get a value of 2.5773g H₂0. Since you have 2 sig figs in the question your answer should have 2 sig figs, so you get an answer of 2.6g H₂O.

Solution/ part 2: To solve this part you must convert the grams of CO₂ to moles of CO₂. From there you cross the mole bridge, which is the mole ratio between the molecules. Finally you must convert the moles of glucose to grams of glucose. Then you will plug all of those numbers into your calculator to get your solution.

For this part you are trying to solve for how much glucose will be formed given 6.3g CO₂.

Step 1- The first step to solving this problem is to convert the grams of CO₂ to moles of CO₂. To do this you will divide 1 mol CO₂ by the molar mass of CO₂. You find the molar mass by looking on the periodic table. You then multiply 6.3g CO₂ by the value you found which is \[\frac{1molCO_{2}}{44gCO_{2}}.\]

\[6.3gCO_{2}\frac{1molCO_{2}}{44gCO_{2}}\]

Step 2- Next, you need to cross the mole bridge. This is where you show the mole ratio between the molecules. This simply means that you will be going from moles of CO₂ to moles of C₆H₁₂O₆. You do this because you ae trying to solve for how much glucose will be formed with the reaction and you started with CO_2.The mole ratio of CO₂ to C₆H₁₂O₆ is 6:1. This is given in the equation. Now you need to divide moles of C₆H₁₂O₆ by the moles of CO₂. Which is \[\frac{1molC_{6}H_{12}O_{6}}{6molCO_{2}}\]. Then you take Step 1 and multiply is by this value to give you

\[6.3gCO_{2}\frac{1molCO_{2}}{44gCO_{2}}*\frac{1molC_{6}H_{12}O_{6}}{6molCO_{2}}\]

Step 3- Now you need to convert moles to grams. To do that you need to multiply by the molar mass of C₆H₁₂O₆ which is 180 grams per mole.\[\frac{180gC_{6}H_{12}O_{6}}{1molC_{6}H_{12}O_{6}}\]. Then you multiply this value by step 2 to give you

\[6.3gCO_{2}\frac{1molCO_{2}}{44gCO_{2}}*\frac{1molC_{6}H_{12}O_{6}}{6molCO_{2}}*\frac{180gC_{6}H_{12}O_{6}}{1molC_{6}H_{12}O_{6}}\]

Step 4- Now that you have all your values you just plug them into your calculator and get a value of 4.295g C₆H₁₂O₆. Since you had two sig figs in the question, there will be two sig figs. So your answer will be 4.2g C₆H₁₂O₆ formed.

6.35

Draw all of the possible Lewis structures for the following ions or molecules: include resonance structures and label formal charges on each ion.

1. nitrate ion (NO₃^-)
2. formate ion (CHO₂^-)
3. cyclobutadiene (C₄H₄)
4. ozone (O₃)

The octet rule states that each atom needs to have 8 electrons. Each lone pair counts as 2. A bond also counts as 2 for each atom in the bond. Then you have to look at how many bonds each element likes to have. That is based on their column or period on the periodic table. Carbon likes to have 4 bonds, Oxygen prefers 2, Nitrogen prefers 3, and Chlorine prefers 1. If they have more or less then that it gives them a formal charge. The formula for formal charge is the [number of valance electrons] -[electrons in lone bonding pairs+1/2 the number of bonding electrons]. The Lewis structures will be as follows:

1.) <--> <-->

2.) <-->

3.) <-->

4.) <-->