[marebare]

Hi Everyone,

Im new to this forum, i joined cause i am taking chemistry I right now in college, i did not take chemistry in high school so im really new to all this. i have been doing ok untill this last week. we started doing chemical formulas, and now im just lost.

here is the assignment i have to do, maybe someone could help me?


6CO2 + 6H2O -----> C6H12O6 + 6O2

Q: CALCULATE THE FORMULA WEIGHT FOR EACH OF TH 4 COMPONENTS ( USE WHOLE #'S) DO BOTH SIDES BALANCE?

Q IF A TREE PRODUCES 100 g OF GLUCOSE IN A DAY HOWMANY GRAMS OF CARBON DIOXIDE IS IT USING UP AND HOW AMY GRAMS OF OXYGEN IS IT PRODUCING IN THAT TIME. SHOW CALCULATIONS?

[Arkcon]

Well, lets start with the first one.  You can look up the definition of formula weight online or in your book, then attempt it for the 4 compounds in the reaction and we'll tell you if we see a problem.  Hint: formula mass has nothing to do with balancing a reaction.

[ARGOS++]

Dear Marebare;

For Q1:  You may use the “Periodic Table” and count the “Atom Masses” together.

For Q2:  That’s a normal “Stoichiometry Problem”.
               So you may apply the recipe/scheme in:   "Stoichiometry Problem”

(Sorry! Arkcos,  once again.)

Good Luck!
                    ARGOS⁺⁺

[marebare]

ok so the deffinition is:
the sum the atomic masses of a chemical compound as indicated by the formula expressed in atomic mass units
 so do i just add up all the masses of each atom? like hydrogen has a atomic mass of 1 and if there are 2 hydrogen atoms then the atomic mass in this formula is 2?

[marebare]

ok nevermind i found this:

formula weight in chemistry, a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a formula by the number of atoms of that element present in the formula, and then adding all of these products togethe


so the correct answer would be


6CO2 = 162 amu

[Arkcon]

Yep.  If you're a biochemist it might be worthwhile for you to memorize the atomic weights of common elements in biological systems -- H=1, C=12, O=16, N=14, I look up S and P when I need 'em.

[ARGOS++]

Dear Marebare;

Sorry!,  –   I got  44  =  MW of CO₂ ( = 12 + 2 * 16).

The Factor/Indices 6 you need later, in Q2.

Good Luck!
                    ARGOS⁺⁺

[Arkcon]

so the correct answer would be


6CO2 = 162 amu

No, (12 + 16 +16) *6 = 264.  But like ARGOS++ said, and I did as well here ...

Hint: formula mass has nothing to do with balancing a reaction.

We don't use summed formula masses to balance chemical equations.

[marebare]

ok thanks guys

[marebare]

ok so this is what i did for Q1


6CO2 = (12+15+15)6= 252
6H2O=(1+1+15)6=103

252+103 = 354

C6H1206 (72+12+96) = 180
6HH2O(1+1+15)6=102
180+102 = 282

SO THE ANSWER WOULD BE NO MASS ON EACH SIDE DO NOT BALANCE?

CAN ANYONE HELP ME ON Q2? NOT SURE WERE TO START

[Borek]

Oxygen mass is 16, not 15.

Check your calculation of mass of reagents on the RHS.

[ARGOS++]

Dear Marebare;

For Q1 you have only (and not more) do to the same as I did for CO₂ in the Posting # 6 above, for  H₂O, and C₆H₁₂O₆, and O₂!
That’s all!

For Q2 you will need ALL your 6, as I already told you!
Remember:    
With your Reaction Equation you have already done line B.), and by answering Q1 you have already done line C₁.) of the Scheme in:   "Stoichiometry Problem”


Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

Dear Marebare;

After I have seen (Yes) you have read/study the:   "Stoichiometry Problem”;

And after I have also seen you figured finally out the right values in your parallel Topic:  "Is this right?”, quite in hardest manner (as you told), I will show you how easy you would have finished this Problem, if you would have continued this particular Topic.

(I “quite never” yell at anybody!, especially not in such a case like this, where I realise, that the questioner really try to learn.  And: I fell sad with “each” person who lost another very associated person; I lost a few (too much) myself.)

For Q1:
MW’s:  CO₂  =  12 + 2 * 16  =  44 g/m.
           H₂O  =  2 * 1 + 16   =  18 g/m.
           O₂    =  2 * 16         =  32 g/m.
           C₆H₁₂O₆  =  6 * 12 + 12 * 1 + 6 * 16 = 180 g/m.
That’s already all.

For Q2:
Till yet your Scheme should look with your answers from Q1 like:
A.)    Reaction:         CO₂   +         H₂O    ----->        C₆H₁₂O₆   +         O₂
B.)    Balancing:     6 CO₂   +      6 H₂O    ----->     1  C₆H₁₂O₆   +      6  O₂

C₁.)   MW’s:             44 g/m        18 g/m                     180 g/m          32 g/m
C_m.)  Masses:                                                          100.0 g
C₂.)   Moles:                                                            0.5555 m         

D.)     Multiplier:                                                       0.5555   (because 0.5555 / 1)

Now you have found how many times the Three can, per day, fulfil the Equation in line B.).
So you can copy your only found Multiplier to all other Reactants/Products of your Equation, and that gives you:
D.)     Multiplier:        0.555            0.555                      0.5555            0.555

With the help of the Factors/Indices of line B.) you can now calculate upwards:
C₂.)   Moles:       6 * 0.555     6 * 0.555                       0.5555 m    6 * 0.555
And that results in:
C₂.)   Moles:           3.333 m      3.333 m                       0.5555 m         3.333 m

And finally you convert this Moles back into Grams with the MW’s of line B.), and that gives you the complete line C_m.):   
C_m.)  Masses:          146.6 g           60.0                      100.0 g           106.6 g

Now you know/read that your Three will “convert”, per day, 146.6 g CO₂ and 60 g  H₂O   into  100.0 g Glucose and 106.6 g  Oxygen.

You can validate your calculation on line C_m.), because 146.6 + 60.0  =  100.0 + 106.6, and you see:  You are perfect!


Good Luck!
                    ARGOS⁺⁺

P.S.:   Marebare:  I had already pre-prepared this final Replay for you before I realised your “Parallel Topic”.  Of course, a little more detailed, but that’s now not required anymore. I hope it will be anyway of some help to you for future similar Problems.