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2. How many atoms, molecules, or formula units of the substance are in the following amounts? a) 3.50 mol of O b) 2.75 g of Se c) 5.50 g of CH22O11 d) 5.00 g of Al(NO3) e) 5.00 ml of ethanol, CH₂CH₂OH, density = 0.790 g/ml f) a sphere of chromium metal, 0.343 mm in diameter. The density of chromium is 7.20 g/cm³ solve everything step by step, cancellation by cancellation

Accepted Answer

Here are the step-by-step solutions to calculate the number of atoms, molecules, or formula units in each case:
a) 3.50 mol of O

Conversion factor: 1 mol O = 6.022 x 10²³ atoms O (Avogadro's number)

Calculation:
3.50 mol O x (6.022 x 10²³ atoms O / 1 mol O) = 2.11 x 10²⁴ atoms O
b) 2.75 g of Se

Conversion factors:

1 mol Se = 78.96 g Se (molar mass of Se)

1 mol Se = 6.022 x 10²³ atoms Se (Avogadro's number)

Calculation:
2.75 g Se x (1 mol Se / 78.96 g Se) x (6.022 x 10²³ atoms Se / 1 mol Se) = 2.10 x 10²² atoms Se
c) 5.50 g of CH₂₂O₁₁ (sucrose)

Conversion factors:

1 mol CH₂₂O₁₁ = 342.30 g CH₂₂O₁₁ (molar mass of sucrose)

1 mol CH₂₂O₁₁ = 6.022 x 10²³ molecules CH₂₂O₁₁ (Avogadro's number)

Calculation:
5.50 g CH₂₂O₁₁ x (1 mol CH₂₂O₁₁ / 342.30 g CH₂₂O₁₁) x (6.022 x 10²³ molecules CH₂₂O₁₁ / 1 mol CH₂₂O₁₁) = 9.68 x 10²¹ molecules CH₂₂O₁₁
d) 5.00 g of Al(NO₃)₃

Conversion factors:

1 mol Al(NO₃)₃ = 213.00 g Al(NO₃)₃ (molar mass of Al(NO₃)₃)

1 mol Al(NO₃)₃ = 6.022 x 10²³ formula units Al(NO₃)₃ (Avogadro's number)

Calculation:
5.00 g Al(NO₃)₃ x (1 mol Al(NO₃)₃ / 213.00 g Al(NO₃)₃) x (6.022 x 10²³ formula units Al(NO₃)₃ / 1 mol Al(NO₃)₃) = 1.41 x 10²² formula units Al(NO₃)₃
e) 5.00 mL of ethanol, CH₂CH₂OH, density = 0.790 g/mL

Conversion factors:

1 mL ethanol = 0.790 g ethanol (density)

1 mol CH₂CH₂OH = 46.07 g CH₂CH₂OH (molar mass of ethanol)

1 mol CH₂CH₂OH = 6.022 x 10²³ molecules CH₂CH₂OH (Avogadro's number)

Calculation:
5.00 mL ethanol x (0.790 g ethanol / 1 mL ethanol) x (1 mol CH₂CH₂OH / 46.07 g CH₂CH₂OH) x (6.022 x 10²³ molecules CH₂CH₂OH / 1 mol CH₂CH₂OH) = 6.51 x 10²² molecules CH₂CH₂OH
f) a sphere of chromium metal, 0.343 mm in diameter, density = 7.20 g/cm³

Conversion factors:

1 mm = 0.1 cm (conversion between millimeters and centimeters)

1 cm³ = 1 mL (conversion between cubic centimeters and milliliters)

1 mL = 1 g (density of chromium)

1 mol Cr = 52.00 g Cr (molar mass of chromium)

1 mol Cr = 6.022 x 10²³ atoms Cr (Avogadro's number)

Calculation:
1. Calculate the volume of the sphere:

Volume = (4/3)πr³, where r is the radius (diameter/2)
Volume = (4/3)π(0.343 mm / 2)² = 0.0212 cm³
2. Calculate the mass of the sphere:

Mass = Volume x Density
Mass = 0.0212 cm³ x 7.20 g/cm³ = 0.153 g Cr
3. Calculate the number of chromium atoms:
0.153 g Cr x (1 mol Cr / 52.00 g Cr) x (6.022 x 10²³ atoms Cr / 1 mol Cr) = 1.77 x 10²¹ atoms Cr


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