6. How many grams of CaCO 3 are present in a sample if there are 4.52 x 10 24 at

Question

6. How many grams of CaCO3 are present in a
sample if there are 4.52 x 1024 atoms of carbon in that sample?

7. The empirical formula of an unknown substance is HSO4.
If the molecular weight of the substance 194 g/mol, what is its molecular
formula?

8. The empirical formula of an unknown substance is
determined to be CH2. What is its molecular formula if the
molecular weight is 84 g/mole?

9. Nicotine has an empirical formula of C5H7N
and a molecular weight of 162 g/mol. What is its molecular formula?

10.. How many moles are in the following:

a. 1.29
x 1024 hydrogen atoms in HF

b. 7.36
x 1024 free oxygen atoms

c. 3.28 x 1023 Na atoms in salt (NaCl)

11. How many atoms are present in the following?

a. Hydrogen
atoms in 3.14 moles of H2O

b. Nitrogen
atoms in 2.79 moles of NH3

c. Oxygen
atoms in 4.90 moles in Al2O3

12. How many atoms are in a 3.56 g sample of Cu? Is this a
mole of Cu?

Explanation / Answer

6. moles of Carbon = 4.52*10^24/(6.02*10^23) = 7.5 moles.

So, moles of CaCO3 = 7.5

So mass = 100*7.5 = 750g.

7.empirical mass of HSO4 = 97 g

mol. mass = 194 g/mol.

So n= mol. mass/empirical mass = 194/97 =2.

So mol. formula = (HSO4)2 or H2S2O8

8.empirical mass of CH2 = 14 g

mol. mass = 84 g/mol.

So n= mol. mass/empirical mass = 84/14 =6.

So mol. formula = (CH2)6 or C6H12.

9.empirical mass of C5H7N = 81 g

mol. mass = 162 g/mol.

So n= mol. mass/empirical mass = 162/81 =2.

So mol. formula = (C5H7N)2 or C10H14N2.

10. a)moles of HF = 1.29*10^24/(6.02*10^23) = 2.14 moles.

b) moles of O = 7.36*10^24/(6.02*10^23) = 12.226 moles.

c)moles of Na = 3.28*10^23/(6.02*10^23) = 0.545 moles.

11.a) moles of H = 2* 3.14 = 6.28

So atoms = 6.28*6.022*10^23 = 3.78*10^24.

b)moles of N = 2.79

So atoms = 2.79*6.022*10^23 = 1.68*10^24

c)moles of O = 3*4.9 = 14.7

So atoms = 14.7*6.022*10^23 = 8.85*10^24 atoms

12. moles of Cu = 3.56/63.5 = 0.056

So atoms = 0.056*6.022*10^23 = 0.3376*10^23 atoms

So its not a mole.

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