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**DATA:**

Number of atoms= 7.5*10^{15}

Mass in grmas=?

**SOLUTION:**

6.02*10^{23} atoms are
present in 1 mol of Ni

7.5*10^{15} atoms are
present in (1*7.5*10^{15})/6.02*10^{23}

= 1.2458*10^{-8} mol

Mole of Ni = mass of Ni in grmas/ atomic mass of Ni

Mass of Ni= mole of Ni * atomic mass of Ni

Mass of Ni = 1.2458*10^{-8}
* 58.6934

**Mass of Ni = 7.3123*10 ^{-7}g**

**DATA: **

Mass of aspirin= 0.1g

Number of molecules = ?

SOLUTION:

Mole of aspirin = mass of aspirin/ molecular mass of aspirin

= 0.1 /180

= 5.5555*10^{-4}

1 mol aspirin contains 6.02*10^{23
}molecules

5.5555*10^{-4} contains (6.02*10^{23}*
5.5555*10^{-4})

**
=3.3444*10 ^{20} molecules**

Please let me know if you need any clarification. I'm always happy to answer your questions.

i am getting back to you with the solution of other three in a while. hope you'll understand

DATA:

Percentage of Ag= 63.5%

Percentage of N= 8.25%

Percentage of O_{2} =
100-63.5-8.25= 28.25%

Empirical formula=?

**SOLUTION**

**STEP 1:**

DETERMINE MOLES

Mol of Ag = 63.5/107.86 = 0.5887mol

Mole of N = 8.25/14.00 = 0.5892mol

Mole of O_{2} =
28.25/15.99 = 1.7667mol

**STEP 2:**

DTERMINE RATIOS

Ag= 0.5887/0.5887 = 1

N= 0.5892/0.5887 = 1.000

O_{2} = 1.7667/0.5887 =
3.00

**STEP 3:**

**WRITW EMPIRICAL FORMULA**

**AgNO _{3}**

**DATA:**

Molecular mass = 42g/mol

Empirical formula = CH_{2}

Molecular formula =?

SOLUTION:

**STEP 1:**

DETERMINE ānā

n = molecular mass/empirical formula mass

n= 42/14

**n= 3**

**STEP 2**

DTERMINE MOLECULAR FORMULA

Molecular formula = (empirical
formula)_{n}

Molecular formula = (CH_{2})_{3}

**Molecular formula = C _{3}H_{6}**

**DATA:**

Mass of sample= 60g

Percentage of Pb = 64%

Percentage of C = 29.7%

Percentage of H = 6.2%

Empirical formula=?

**SOLUTION:**

**STEP 1**

DETERMINE MASS OF ELEMENTS IN THE GIVEN MASS OF SAMPLE

Mass of Pb= (64*60)/100 = 38.4g

Mass of C= (29.7*60)/100 = 17.82g

Mass of H = (6.2*60)/100 = 3.72g

**STEP 2**

DETERMINE MOLES

Moles of Pb = 38.4/207.2 = 0.1853mol

Moles of C = 17.82/12.01= 1.4837mol

Moles of H= 3.72/1.00 = 3.72mol

**STEP 3**

DETERMINE RATIOS

Pb= 0.1853/0.1853 = 1

C= 1.4837/0.1853= 8.00

H= 3.72/0.1853= 20.07

**STEP 4**

WRITE EMPIRICAL FORMULA

**C _{8}H_{20}Pb**

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