Chemistry Grams to Moles to Molecules to Moles to Grams Conversion ... - Free Printable

To solve this problem, we need to convert the number of particles (atoms or molecules) into moles.

The Rule:
To go from Particles → Moles, you divide by Avogadro's number.
Avogadro's number is approximately $6.022 \times 10^{23}$.

The Formula:
$$ \text{Moles} = \frac{\text{Number of Particles}}{6.022 \times 10^{23}} $$

Here are the step-by-step calculations for each part:

Part 1: Convert Particles in Moles to Grams

*(Note: The header says "Particles to Grams", but the instructions say "Convert the following from Particles to Moles". Based on the units given ($1.50 \times 10^{23}$ atoms), these are particle counts. We will calculate the Moles as requested by the specific instruction line.)*

1. $1.50 \times 10^{23}$ atoms of F₂
* Calculation: $\frac{1.50 \times 10^{23}}{6.022 \times 10^{23}}$
* The $10^{23}$ cancels out.
* $1.50 / 6.022 \approx 0.249$
* Answer: $0.249$ mol F₂

2. $1.50 \times 10^{23}$ molecules of H₂O
* Calculation: $\frac{1.50 \times 10^{23}}{6.022 \times 10^{23}}$
* Same as above.
* Answer: $0.249$ mol H₂O

3. $2.79 \times 10^{24}$ atoms of Al
* Calculation: $\frac{2.79 \times 10^{24}}{6.022 \times 10^{23}}$
* Handle powers of 10: $10^{24} / 10^{23} = 10^1 = 10$.
* Numbers: $2.79 / 6.022 \approx 0.4633$.
* Multiply by 10: $0.4633 \times 10 = 4.633$.
* Answer: $4.63$ mol Al

4. $1.15 \times 10^{23}$ molecules of CO₂
* Calculation: $\frac{1.15 \times 10^{23}}{6.022 \times 10^{23}}$
* $1.15 / 6.022 \approx 0.1909$
* Answer: $0.191$ mol CO₂

5. $5.25 \times 10^{25}$ atoms of Pb
* Calculation: $\frac{5.25 \times 10^{25}}{6.022 \times 10^{23}}$
* Handle powers of 10: $10^{25} / 10^{23} = 10^2 = 100$.
* Numbers: $5.25 / 6.022 \approx 0.8718$.
* Multiply by 100: $87.18$.
* Answer: $87.2$ mol Pb

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Part 2: Convert Particles in Moles to Grams

*(Again, calculating Moles from the given particle counts)*

1. $4.45 \times 10^{24}$ molecules of N₂
* Calculation: $\frac{4.45 \times 10^{24}}{6.022 \times 10^{23}}$
* Powers: $10^1 = 10$.
* Numbers: $4.45 / 6.022 \approx 0.7389$.
* Result: $7.389$.
* Answer: $7.39$ mol N₂

2. $8.91 \times 10^{21}$ atoms of Zn
* Calculation: $\frac{8.91 \times 10^{21}}{6.022 \times 10^{23}}$
* Powers: $10^{21} / 10^{23} = 10^{-2} = 0.01$.
* Numbers: $8.91 / 6.022 \approx 1.479$.
* Result: $1.479 \times 0.01 = 0.01479$.
* Answer: $0.0148$ mol Zn

3. $1.05 \times 10^{24}$ atoms of S
* Calculation: $\frac{1.05 \times 10^{24}}{6.022 \times 10^{23}}$
* Powers: $10^1 = 10$.
* Numbers: $1.05 / 6.022 \approx 0.1743$.
* Result: $1.743$.
* Answer: $1.74$ mol S

4. $1.05 \times 10^{24}$ atoms of P₄
* *Note: Even though it is P₄, if the question asks for atoms of P₄, we treat the count as the entity being divided. If it meant molecules of P, the math is the same for finding moles of that entity.*
* Calculation: $\frac{1.05 \times 10^{24}}{6.022 \times 10^{23}}$
* Same as #3.
* Answer: $1.74$ mol P₄

5. $2.75 \times 10^{22}$ molecules of O₂
* Calculation: $\frac{2.75 \times 10^{22}}{6.022 \times 10^{23}}$
* Powers: $10^{-1} = 0.1$.
* Numbers: $2.75 / 6.022 \approx 0.4566$.
* Result: $0.04566$.
* Answer: $0.0457$ mol O₂

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Part 3: Convert particles below from Particles to Moles

1. $5.34 \times 10^{24}$ molecules of CH₄
* Calculation: $\frac{5.34 \times 10^{24}}{6.022 \times 10^{23}}$
* Powers: $10$.
* Numbers: $5.34 / 6.022 \approx 0.8867$.
* Result: $8.867$.
* Answer: $8.87$ mol CH₄

2. $2.87 \times 10^{23}$ formula units of NaCl
* Calculation: $\frac{2.87 \times 10^{23}}{6.022 \times 10^{23}}$
* Numbers: $2.87 / 6.022 \approx 0.4765$.
* Answer: $0.477$ mol NaCl

3. $4.87 \times 10^{25}$ molecules of C₆H₁₂O₆
* Calculation: $\frac{4.87 \times 10^{25}}{6.022 \times 10^{23}}$
* Powers: $100$.
* Numbers: $4.87 / 6.022 \approx 0.8087$.
* Result: $80.87$.
* Answer: $80.9$ mol C₆H₁₂O

4. $1.19 \times 10^{23}$ atoms of Ne
* Calculation: $\frac{1.19 \times 10^{23}}{6.022 \times 10^{23}}$
* Numbers: $1.19 / 6.022 \approx 0.1976$.
* Answer: $0.198$ mol Ne

5. $7.92 \times 10^{22}$ molecules of I₂
* Calculation: $\frac{7.92 \times 10^{22}}{6.022 \times 10^{23}}$
* Powers: $0.1$.
* Numbers: $7.92 / 6.022 \approx 1.315$.
* Result: $0.1315$.
* Answer: $0.132$ mol I₂

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Part 4: Convert particles below from Moles to Grams

*(Wait, the header says "Moles to Grams" but the data is clearly particles again: "$5.34 \times 10^{24}$ atoms". The instruction says "Convert... from Moles to Grams" but the input is particles. Given the pattern of the previous sections, it is highly likely the student is expected to convert Particles to Moles again, despite the confusing header. However, looking at Part 5 and 6, they explicitly ask for grams. Let's look closely at Part 4's instruction line: "Convert particles below from Moles to Grams." This is contradictory. Usually, these worksheets progress: Part 1-3 are Particles->Moles. Part 4-6 might be Particles->Grams. Let's calculate Particles to Grams for Part 4, 5, and 6, as that is the standard progression for these "Work Ahead" sheets.)*

How to convert Particles to Grams:
1. Divide particles by $6.022 \times 10^{23}$ to get Moles.
2. Multiply Moles by the Molar Mass (from periodic table) to get Grams.

Molar Masses needed:
* Ca: $40.08$ g/mol
* Cu: $63.55$ g/mol
* Br₂: $2 \times 79.90 = 159.8$ g/mol
* He: $4.00$ g/mol
* Mg(OH)₂: $24.31 + 2(16.00 + 1.01) = 58.33$ g/mol

Part 4 Calculations (Particles → Grams):

1. $5.34 \times 10^{24}$ atoms of Ca
* Moles: $\frac{5.34 \times 10^{24}}{6.022 \times 10^{23}} = 8.867$ mol
* Grams: $8.867 \text{ mol} \times 40.08 \text{ g/mol} = 355.4$ g
* Answer: $355$ g Ca

2. $2.87 \times 10^{23}$ atoms of Cu
* Moles: $\frac{2.87 \times 10^{23}}{6.022 \times 10^{23}} = 0.4766$ mol
* Grams: $0.4766 \text{ mol} \times 63.55 \text{ g/mol} = 30.29$ g
* Answer: $30.3$ g Cu

3. $4.87 \times 10^{25}$ molecules of Br₂
* Moles: $\frac{4.87 \times 10^{25}}{6.022 \times 10^{23}} = 80.87$ mol
* Grams: $80.87 \text{ mol} \times 159.8 \text{ g/mol} = 12,923$ g
* Answer: $1.29 \times 10^4$ g Br₂ (or $12,900$ g)

4. $1.19 \times 10^{23}$ atoms of He
* Moles: $\frac{1.19 \times 10^{23}}{6.022 \times 10^{23}} = 0.1976$ mol
* Grams: $0.1976 \text{ mol} \times 4.00 \text{ g/mol} = 0.790$ g
* Answer: $0.790$ g He

5. $7.92 \times 10^{22}$ formula units of Mg(OH)₂
* Moles: $\frac{7.92 \times 10^{22}}{6.022 \times 10^{23}} = 0.1315$ mol
* Grams: $0.1315 \text{ mol} \times 58.33 \text{ g/mol} = 7.67$ g
* Answer: $7.67$ g Mg(OH)₂

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Part 5: Convert particles below from Particles to Moles

*(The header says "Particles to Moles" but the instruction line says "Convert... from Moles to Grams". Looking at the values, they are particles. Given Part 4 was likely Particles->Grams, let's re-read carefully. Actually, Part 4 header was "Moles to Grams" but inputs were particles. Part 5 header is "Particles to Moles". Instruction says "Convert... from Moles to Grams". This worksheet has typos. Let's look at the magnitude. If we just do Particles -> Moles, the numbers are small. If we do Particles -> Grams, they are larger. Let's assume the Instruction Line is the correct task for the specific row, but the Header describes the section. Wait, Part 5 Header: "Particles to Moles". Instruction: "Convert... from Moles to Grams". This is very confusing. Let's look at Part 6. Header: "Particles to Grams". Instruction: "Convert... from Moles to Grams".*

*Let's stick to the most logical educational path:*
*Parts 1-3:* Particles → Moles.
*Parts 4-6:* Particles → Grams. (Since converting particles directly to moles was already practiced extensively).

However, looking at Part 5, Question 1: $1.19 \times 10^{23}$ atoms of Li.
If Particles → Moles: $0.198$ mol.
If Particles → Grams: $0.198 \times 6.94 = 1.37$ g.

Let's look at the provided text in the image for Part 5 again.
"Part 5: Convert particles below from Particles to Moles."
"Convert the following from Moles to Grams."
This is a direct contradiction. Usually, the bold header is the section topic. The italicized instruction might be a copy-paste error from the previous section.
BUT, Part 6 Header is "Particles to Grams".
If Part 5 is "Particles to Moles", why would Part 6 exist?
Let's assume Part 5 is Particles → Moles (following the header) and Part 6 is Particles → Grams (following the header). The italicized instructions seem to be erroneous copies.

Re-evaluating based on Headers:
* Part 1 Header: Particles to Grams (Instruction: Particles to Moles). We did Particles to Moles.
* Part 2 Header: Particles to Grams (Instruction: Particles to Moles). We did Particles to Moles.
* Part 3 Header: Particles to Moles (Instruction: Particles to Moles). We did Particles to Moles.
* Part 4 Header: Moles to Grams (Instruction: Moles to Grams). Inputs are Particles. We did Particles to Grams.
* Part 5 Header: Particles to Moles. Instruction: Moles to Grams. Inputs are Particles.
* If we follow the header, we do Particles → Moles.
* If we follow the instruction, we do Particles → Grams.
* Given Part 4 was likely Grams, and Part 6 is definitely Grams, Part 5 being Moles makes sense as a mix-up or intermediate step. However, usually, these sheets group them. Let's look at the numbers.
* Let's provide Particles to Moles for Part 5 because the Header explicitly says "Particles to Moles", which distinguishes it from Part 6 "Particles to Grams".

Part 5 Calculations (Particles → Moles):

1. $1.19 \times 10^{23}$ atoms of Li
* Calculation: $\frac{1.19 \times 10^{23}}{6.022 \times 10^{23}}$
* Answer: $0.198$ mol Li

2. $5.34 \times 10^{24}$ molecules of CO₂
* Calculation: $\frac{5.34 \times 10^{24}}{6.022 \times 10^{23}} = 8.867$
* Answer: $8.87$ mol CO₂

3. $2.87 \times 10^{23}$ atoms of Ag
* Calculation: $\frac{2.87 \times 10^{23}}{6.022 \times 10^{23}} = 0.4766$
* Answer: $0.477$ mol Ag

4. $4.87 \times 10^{25}$ molecules of C₆H₁₂O₆
* Calculation: $\frac{4.87 \times 10^{25}}{6.022 \times 10^{23}} = 80.87$
* Answer: $80.9$ mol C₆H₁₂O₆

5. $7.92 \times 10^{22}$ formula units of Mg(OH)₂
* Calculation: $\frac{7.92 \times 10^{22}}{6.022 \times 10^{23}} = 0.1315$
* Answer: $0.132$ mol Mg(OH)₂

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Part 6: Convert particles below from Particles to Grams

*(Header and Instruction agree: Convert to Grams. Inputs are Particles.)*

Molar Masses needed:
* Ca: $40.08$ g/mol
* Cu: $63.55$ g/mol
* Br₂: $159.8$ g/mol
* He: $4.00$ g/mol
* Mg(OH)₂: $58.33$ g/mol

1. $1.19 \times 10^{23}$ atoms of Ca
* Moles: $0.1976$ mol (from Part 5 calc style)
* Grams: $0.1976 \times 40.08 = 7.92$ g
* Answer: $7.92$ g Ca

2. $5.34 \times 10^{24}$ molecules of Cu
* *Note: Copper is an element, so it exists as atoms, not molecules, but we treat the count as the entity.*
* Moles: $8.867$ mol
* Grams: $8.867 \times 63.55 = 563.5$ g
* Answer: $564$ g Cu

3. $2.87 \times 10^{23}$ atoms of Br₂
* *Note: Bromine is diatomic. If these are atoms of Br, we use atomic mass. If "atoms of Br₂" implies molecules, we use molecular mass. Standard chemistry notation "atoms of [Element]" usually means atomic mass. But Br₂ is a molecule. Let's assume the question means "particles of Br₂" (molecules) because it writes Br₂. If it meant atoms, it would say Br. Let's assume Molecules of Br₂.*
* Moles: $0.4766$ mol
* Grams: $0.4766 \times 159.8 = 76.16$ g
* Answer: $76.2$ g Br₂

4. $4.87 \times 10^{25}$ molecules of He
* *Note: Helium is monatomic. "Molecules" here just means particles.*
* Moles: $80.87$ mol
* Grams: $80.87 \times 4.00 = 323.5$ g
* Answer: $324$ g He

5. $7.92 \times 10^{22}$ formula units of Mg(OH)₂
* Moles: $0.1315$ mol
* Grams: $0.1315 \times 58.33 = 7.67$ g
* Answer: $7.67$ g Mg(OH)₂

Final Answer:
Part 1 (Particles to Moles)
1. 0.249 mol F₂
2. 0.249 mol H₂O
3. 4.63 mol Al
4. 0.191 mol CO₂
5. 87.2 mol Pb

Part 2 (Particles to Moles)
1. 7.39 mol N₂
2. 0.0148 mol Zn
3. 1.74 mol S
4. 1.74 mol P₄
5. 0.0457 mol O₂

Part 3 (Particles to Moles)
1. 8.87 mol CH₄
2. 0.477 mol NaCl
3. 80.9 mol C₆H₁₂O₆
4. 0.198 mol Ne
5. 0.132 mol I₂

Part 4 (Particles to Grams)
1. 355 g Ca
2. 30.3 g Cu
3. 1.29 × 10⁴ g Br₂ (or 12,900 g)
4. 0.790 g He
5. 7.67 g Mg(OH)₂

Part 5 (Particles to Moles)
1. 0.198 mol Li
2. 8.87 mol CO₂
3. 0.477 mol Ag
4. 80.9 mol C₆H₁₂O₆
5. 0.132 mol Mg(OH)₂

Part 6 (Particles to Grams)
1. 7.92 g Ca
2. 564 g Cu
3. 76.2 g Br₂
4. 324 g He
5. 7.67 g Mg(OH)₂