Chemistry question
How many atoms of hydrogen are contained in each of the following? 1) 48 molecules of CH3CH2COOH 2) 4.54 mol H3PO4 3) 30 g C6H5ONH
Q1: Suppose the question would be for 1 molecule can you calculate the # H atoms? Q2: Do you know how many atoms are in 1 mol? Q3: Can you convert from grams to moles?
@LeibyStrauss I am confused because I missed the class.
I'll try to help, but I can't replace a whole class in these short posts. I'll reference each post with Q1, Q2, Q3 when replying please do the same. For Q1: \[CH _{3}CH _{2}COOH\] Let's start with something simpler. Do you know how many H's (hydrogen's are in \[H _{2}O\]
2?
@LeibyStrauss
So how many H atoms are in CH3CH2COOH?
6?
Correct. This though, is just for one molecule, it needs to be multiplied by the # of molecules in the example.
Q2: Look in you textbook or google, how many atoms are in a mole or look / search for Avogadro's numv\ber. A mol btw is is just a quantity number, just like a dozen is just the number 12. A dozen eggs and a dozen doughnuts, both mean that you have 12. The mol is same thing.
So it is 6.022 x 10^23
Re: Q1 are you good with the first question?
Yes. Question one is 288
Re: Q2 You are correct. I'm blacking out for a moment. I think 1 mol of H3PO4 would be (since there are 3 H's) 3*6.022*10^23. Since there are 4.54 mol it should be (3)(4.54)(6.022)*10^23. Can someone verify that this is correct?
You are correct re: Q1
Re: Q3. You need to convert from grams to moles and do the same as you did for Q2. Do you know how to convert from grams to moles?
Not really.
Ok. In short it's called dimensional analysis KhanAcademy.org has good videos on it and practice skills. You need to start with whats given, in this case 30 g C6H5ONH. Then set up your equation so that grams cancel each other out and you are left with moles. You need to calculate the molar mass, also know as g/mol, or grams per mole of C6H5ONH and plug the gram value in the denominator (where grams is written). The molar masses are on the periodic table \[30gram C6H5ONH * \frac{1 mole }{grams }\] Following this equation will give you the # of moles
30/108.10 is .27750
It looks correct. I only used 2 sig figs so I got 108. 0.27750 is the # of moles. Find out if your professor / teacher is pick with significant figures. If yes, then since 30 is 2 sig figs you need to round to 0.28. Did what I post regarding the 2nd question make sense?
30(14)(6.022 x 10^23) / 0.28 = 9.03e+26
I'm not sure why your using 30(14). In C6H5ONH there are 6 H's. So I think we have: For 1 mole: 6(6.022)x10^23 Since there are 0.28 mol 0.28(6)(6.022)x10^23. My brain has over-trained tonight and I may be making a mistake.
I got it. Thank you so much. I understood perfectly. :)
You're welcome.
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