What is the mass (in grams) of 9.15 * 1024 molecules of methanol (ch3oh) ?

Solution:

first we have to calculate the mass of 1 mole of CH3OH?

C = 12

H3 = 3x1 = 3

O = 16

H = 1

total mass 12+3+16+1=32g.

Total 32g So we know 6.02x10^23 molecules weigh in32g

therefore,

1 molecule weighs 32/6.02x10^23 g

now,

9.95x10^24 molecules weighs 9.95x10^24 x (32/6.02x10^23) = 9.95x32x10/6.02=99.5x32/6.02

=528.90g

thus the mass in grams is 528.90g.

486.68 g

Solution:

Data Given:

Number of Molecules = 9.15 * 10²⁴

M. Mass of Methanol = 32.04 g. mol⁻¹

Mass of Methanol = ?

Step 1: Calculate Moles of Methanol,

Moles = Number of Molecules : 6.022 * 10²³ Molecules. mol⁻¹

Putting value,

Moles = 9.15 * 10²⁴ Molecules : 6.022 * 10²³ Molecules. mol⁻¹

Moles = 15.19 mol

Step 2: Calculate Mass of Methanol:

Moles = Mass : M. Mass

Solving for Mass,

Mass = Moles * M. Mass

Putting values,

Mass = 15.19 mol * 32.04 g. mol⁻¹

Mass = 486.68 g