Question

is there someone online right now that is good with chemistry? please help me, i need to message what im having trouble with. :(

Answer 1

its this thing, i dont know how to fill out the table at all.

Answer 2

For question 1.... I'm working on 2...
when it comes to solving stoichiometric problems, you need to make sure you stay organized and label everything so you can keep track of where you are and what you’re working with. Units matter and they leave you with the correct answer.

I’m not going to do the actual calculations to find the actual number, you can do that, but i’ll give you a blueprint on how to solve almost any and all stoichiometric problems.

Also. Get out your periodic table. You need it when doing stoichiometry.


When looking at the periodic table you’ll find the atomic mass of the element. The units for atomic mass are grams/mol. For example Carbon’s atomic number (which the the basis of the entire mole system by the way) is 12.011 grams/mol or g/mol for short.

Ok knowing this, let’s look at the first problem.
1a
gold Au, has a mass of 35.12 g.

Look up the g/mol of Au. I found it as 196.967g

Therefore Au = 196.967 g/mol. This means there are 196.967 grams of Au/ 1mol Au. You can also write this as 1molAu/196.967grams Au. the ratio remains the same. 1:196.967

35.12g Au * (1mol Au/196.967g Au) = # mol Au
Notice how the grams cancel to leave you with mol.

1b
This part is simply remembering a number, a very special number. 6.022x10^23. This is the number of atoms in 12g of carbon. This is the basis of the mole system. When we find the moles of any element, we can multiply that number of moles by 6.022x10^23 to find out how many atoms of that substance there are. this is called Avogadro’s number (it sounds confusing and takes a while to grasp, but I just wanted to give you some background) .
The units for Avogadro’s number is atoms/mol

so.. now we have #mol Au from the previous problem. we can use the same conversion factor method we used to find the moles, but this time we’ll use it to find the # of atoms.

# mol Au * (6.022x10^23atoms/1mol) = # atoms
again notice how the moles cancel to leave you with atoms.

Answer 3

Question 2
This is where it get’s kind of tricky, but using the same conversion factor method we’ve been using, it won’t be that bad at all.
2a 1.202g sucrose (C12H22O11).
so here we use the same method of finding moles as before, the only difference is that now we have to multiply and add stuff together.

first find the atomic mass of carbon, hydrogen and oxygen
I got them as
C - 12.011g/mol
H - 1.0079g/mol
O - 15.999 g/mol

now you have to look and see how many carbon atoms are in this compound, how many hydrogen atoms are in this compound, and.. you guessed it, how many oxygen atoms are in this compound.

you can tell how many there are by the subscript next to the element symbol. in this particular compound called sucrose, there are 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms.
now you just multiply the atomic mass of the elements by their subscript or “coefficient” and add them together to figure out how many g/mol are in sucrose (C12H22O11)

2a.
(12.011g/molC * 12) + (1.0079g/molH * 22) + (15.999g/molO * 11) = # g/mol sucrose

now do the same thing you did to find moles in the first problem

1.202g sucrose * (1mol/#g sucrose) = #mol sucrose

This is where it gets kind of tricky.
Keep in mind that there is a special relationship between these atoms. It’s a ratio of one atom to every other one in the compound. Meaning, in order for sugar to be, well, sugar… there has to be 12 carbon atoms for every 22 hydrogen atoms for every 11 oxygen atoms bonded together. this will be key in finding part b of problem 2.

so we now have the moles of sucrose. The coefficient also represent the mole ratio between the elements in the compound. like I said before there has to be 12 carbon atoms for every 22 hydrogen atoms for every 11 oxygen atoms bonded together. meaning for every 1 mol of sucrose, there are 12 mol carbon, 22 mol hydrogen, 11 moles oxygen.

so we can use this information to find out how many moles of carbon, how many moles of hydrogen and how many moles of oxygen are in this particular sample of sucrose.

2b
#mol C12H22O11* (12 mol C /1mol C12H22O11) = #mol C
(notice how the molC12H22O11 cancel and leave you with # mol C )

do the same thing for hydrogen and oxygen

2c
This is the same thing we did for part b of question 1.

#mol C * (6.022x10^23 atoms.mol) = #atoms C

etc.

Hope this helped a bunch.

Answer 4

thank you so much, really :)

Answer 5

All good. Do a bunch of problems and it'll become habit.