Question

1. A sample of gold (Au) has a mass of 35.12 g.
a. Calculate the number of moles of gold (Au) in the sample. .

b. Calculate the number of atoms of gold (Au) in the sample

Answer 1

please in order to calculate number of moles, you have to divide your mass, by atomic mass of Au, please keep in mind that atomic mass of Au =197, so?

Answer 2

moles of Au=\[\frac{ 35.12 }{ 197 }=...\]
please continue
@denii_114

Answer 3

Would it be 0.178??

Answer 4

@Michele_Laino

Answer 5

that's right!

Answer 6

please try, it is a constant whose value is very huge

Answer 7

I lost the connection, I repeat:
how many atoms of Au there are in a mole of Au?
that number is very huge, its name is Avogadro constant

Answer 8

Oh.... Um 79?

Answer 9

pratically to answer to your second question, you have to multiply numbers of your moles of Au by the Avogadro constant.
Please do you know what is the value of Avogadro constant?

Answer 10

No, not really...

Answer 11

Avogadro constant is:
\[N _{A}=6.02*10^{23}\]

Answer 12

so yo\[N _{A}*0.178=6.023*10^{23}*0.178=...\]u have to calculate this:
and you will get your number of atoms of Au, please try!

Answer 13

Okay, Please excuse me if im wrong, im tryin.... Wld it be... 1.072??

Answer 14

I was absent the day the we're showing it... Doing the best i can.

Answer 15

your answer is :
\[1.072*10^{23}\]
atoms of Au

Answer 16

Oh Okay.... So from the answer i get, i always add in the \[10^{23}\]?

Answer 17

yes! you have to multiply your answer 1.072 by the exponential factor, don't forget it, please!

Answer 18

Ohhh.. Okay, got it. Thank You so much!! @Michele_Laino

Answer 19

Thank you!

Answer 20

Okay with that i cld answer the rest of the question similar to it. You think you can heklp me with one more thing, please?? @Michele_Laino

Answer 21

@denii_114 yes!

Answer 22

Okay Thank You!! Here it is....
2. A sample of table sugar (sucrose, C12H22O11) has a mass of 1.202 g.
a. Calculate the number of moles of C12H22O11 contained in the sample

b. Calculate the moles of each element in C12H22O11.
Show your work.

c. Calculate the number of atoms of each type in C12H22O11. Show your work.
@Michele_Laino

Answer 23

please you have to calculate the mass of sucrose molecule. now a sucrose molecule
has
12 aoms of C,
22 atoms of H,
and
11 atoms of O

Answer 24

Got it!

Answer 25

so we have to consider tha atomic mass of C, H and O, please take your periodic table
and write the atomic mass of C, H and O

Answer 26

from mine, I read:
atomic mass of C=12
please continue

Answer 27

H=22
O=11
?

Answer 28

please you have to read atomic masses from periodic table of elements, do you have it, on your chemistry textbook?

Answer 29

I'll look online for one...

Answer 30

ok!
atomic mass of H=1.008, which can be rounded to 1

Answer 31

H=1
o=15

Answer 32

please you have to round off, we have O=15.99, namely 16. are you agree?

Answer 33

Yes.

Answer 34

Forgot to round sorry

Answer 35

ok! Now in order to find the mass of sucrose molecule, you have to do this calculus:
\[12*2+22*1+16*11\]

Answer 36

sorry I have made an error, I rewrite:
\[12*12+22*1+11*16=...\]

Answer 37

222, right?

Answer 38

Wait. did sumthing wrong

Answer 39

342

Answer 40

perfect!

Answer 41

in order to find the sucrose moles, you have to do like previous exercise, namely you have to divide the mass of sucrose, by the molecule mass of sucrose:
\[\frac{ 1.202 }{ 342 }=...\]

Answer 42

please try?

Answer 43

do i round too??

Answer 44

i am trying...

Answer 45

Im doing it... this stupid keyboard takes forever to write in.....
I got 0.003.... round? 0.004.

Answer 46

are you sure? please retry

Answer 47

sorry I have made an error, you are right!

Answer 48

yeaa...

Answer 49

ok! now you know that one sucrose mole contains 6.02*10^23 sucrose molecules, but we have only 0.004 molecules , namely 4*10^-3, so we have to multiply 0.004 by Avogadro constant, like the previous exercise, so:
\[0.004*6.02*10^{23}=...\]

Answer 50

So how do you find the moles for each element?? Same as first question?

Answer 51

Oh okay....

Answer 52

2.41

Answer 53

I got 0.0241*10^23

Answer 54

one question... you get the "0.0241"? if it was 2.41

Answer 55

**how you...

Answer 56

why? you have to find 0.004*6.023*10^23

Answer 57

I keep getting 2.40

Answer 58

sorry, please find 0.004*6.023=...

Answer 59

Okay.. NOW i get 0.024...

Answer 60

ok! now you have to multiply 0.024 by 10^23, so you should get:
\[0.024*10^{23}=2.4*10^{-2}*10^{23}=2.4*10^{-2+23}=2.4*10^{21}\]

Answer 61

Yeaa thts what i did.. i was one step ahead of you...i got tht the first time

Answer 62

ok now we know that we have 2.4*10^21 molecules of sucrose. Please note that in one molecule of sucrose we have 12 atoms of C, so how many atoms of C we have in 2.4 molecules of sucrose?

Answer 63

...sorry in 2.4*10^21 molecules of sucrose...

Answer 64

Do i add them into 2.4 * 10^21?

Answer 65

you have to multiply 2.4*10^21 by 12

Answer 66

What i meant. Okay i got... 2.88

Answer 67

please find 2.4*12...

Answer 68

28.8

Answer 69

don't forget the exponential factor....

Answer 70

2.88

Answer 71

Do i do tht for the rest?? H & O??
H= 2.4
O= 3.6

Answer 72

please, note that we have 28.8 *10^21 atoms of C,
52.8*10^21 atoms of H,
26.4*10^21 atoms of O
are you agree?

Answer 73

Yeah...

Answer 74

that answers to c.

Answer 75

Yup got it...

Answer 76

we have to answer to b.

Answer 77

now, please write the value of the mass of an atom of C, please

Answer 78

12

Answer 79

you are right! but I want to know how many grams!

Answer 80

.100g

Answer 81

It is very simple, you have to multiply 12 by another constant, whose value is 1.66*10^-24, so we have:
\[12*1.66*10^{-24}\]

Answer 82

is it clear?

Answer 83

No, where you get the 1.66??

Answer 84

it is another constant whose name is "Dalton", it is tabulated in all chemical tables.
In general in ordr to find a mass (in grams) of an atom, you have to multiply tha atomic mass of that element by the "Dalton"

Answer 85

oh oh oh..... okay...

Answer 86

for example, atomic mass of Fe is 58.6, then the mass of atoms of Fe, expressed in grams is:
\[58.6*1.66*10^{-24}=97.3*10^-24 grams\]

Answer 87

Oh okay.. so it be 19.92 * 10-24grams??

Answer 88

that's right!

Answer 89

So tht would be for C...
H would be .... 1.66 * 10-24??

Answer 90

now you have to find the mass (in grams) of 28.8*10^21 atoms of C, namely you have to multiply your number by 28.8*10^21

Answer 91

so?

Answer 92

Multiply 19.92 * 28.8 ??

Answer 93

\[19.92*10^{-24}*28.8*10^{21}=....\]

Answer 94

0.57??

Answer 95

that's right! we have 0.57 grams of C

Answer 96

and how many moles of C?

Answer 97

0.47?