Question

What is the molecular weight, density, and how many moles in

1) 1mL of water
2) 0.04 g of benzophenone

Answer 1

1)
`M_(H_2O)=18g/mol; \rho_(H_2O)=0.997g/mL;n=0.055mol`

2)
`M_(benzophenone)=182g/mol ;\rho_(benzophenone)=1.11 g/mL;n_(benzophenone)=2.195x10^(-4)mol`

Step-by-step explanation:

Hello,

In this case, the molar mass is obtained by adding each atom's atomic mass multiplied by the number of atoms in the molecule, the density is searched on databases and the moles by using a dimensional analysis with the molar mass starting by the compound's mass, thus, for each case, we obtain:

1)

`M_(H_2O)=2*m_H+m_O=2*1+16=18g/mol`

`\rho_(H_2O)=0.997g/mL`

`n=1mLH_2O*(0.997gH_2O)/(1mLH_2O)*(1molH_2O)/(18gH_2O) =0.055molH_2O`

2)

`M_(benzophenone)=13*m_C+10*m_H+m_O=12*13+10*1+16=182g/mol`

`\rho_(benzophenone)=1.11 g/mL`

`n=0.04g\ benzophenone*(1mol\ benzophenone)/(182g\ benzophenone) =2.195x10^(-4)mol`

Best regards.

Answer 2

1) Molecular weight: 18g/mol

Density: at room temperature (25ºC) is 0.997g/mL

Moles: 0.55moles

2) Molecular weight: 182g/mol

Density: at room temperature (25ºC) is 1.11g/mL

Moles:
`2.2x10^(-4)moles`

Step-by-step explanation:

1) The molecular formula of water is
`H_(2)O` thus, the molecular weight is the sum of the weights of its atoms.

H: 1g/mol x 2 = 2g/mol

O: 16g/mol x 1 = 16g/mol

`H_(2)O`: 2g/mol + 16g/mol = 18g/mol

The density (δ) of water at room temperature (25ºC) is 0.997g/mL

Therefore, the weight (m) of 1mL of water is:

m = δ.V =
`0.997(g)/(mL). 1mL = 0.997g`

Multiplying the mass by its molecular weight gives the number of moles:

`(0.997g )/(18g/mol) = 0.55 moles`

2) The molecular formula of benzophenone is
`C_(13)H_(10)O` thus, the molecular weight is the sum of the weights of its atoms.

C: 12g/mol x 13 = 156g/mol

H: 1g/mol x 10 = 10g/mol

O: 16g/mol x 1 = 16g/mol

`C_(13)H_(10)O`: 156g/mol + 10g/mol + 16g/mol = 182g/mol

The density (δ) of water at room temperature (25ºC) is 1.11g/mL

Multiplying the mass by its molecular weight gives the number of moles:

`(0.04g )/(182g/mol) = 2.2x10^(-4)moles`