Problems 4-6 convert the measurements into moles. 4. 9.03 x1023 C atoms

5. 2.44x1023 H2O molecules

6. 6.02x1023 Li atoms

4. 1.5moles

5. 0.405moles

6. 1 mol

Explanation:

To find the number of moles in each susbtance in this question, we divide each number of atom/molecule measurement by Avagadro's number (6.02 × 10^23 atoms)

That is; nA = n × 6.02 × 10^23 atoms

n = nA ÷ 6.02 × 10^23 atoms

Where;

nA = number of atoms

n = number of moles

4.) n of Carbon = 9.03 x 10^23 ÷ 6.02 × 10^23

n = 1.5 × 10^ (23-23)

n = 1.5 × 10^0

n = 1.5 × 1

n = 1.5moles.

5.) n of H2O = 2.44 x 10^23 ÷ 6.02 × 10^23

n = 0.405 × 10^ (23-23)

n = 0.405 × 10^0

n = 0.405moles

6.) n of lithium = 6.02 x 10^23 Li atoms ÷ 6.02 × 10^23

n = 1 × 10^ (23-23)

n = 1 × 10^0

n = 1 mol

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.