This article will explain how to calculate the number of sodium (Na+) ions contained in 99.6 mg of sodium sulfite (Na2SO3). Sodium sulfite is an ionic compound composed of sodium cations (Na+) and sulfite anions (SO32-). By using the molar mass of sodium sulfite and stoichiometry, we can determine the moles of Na2SO3 present in 99.6 mg. From the chemical formula, we can then find the number of sodium ions present. This calculation requires using the molar mass and balanced chemical equation for sodium sulfite, along with stoichiometric relationships. The step-by-step working and explanations are provided below.
Step-by-Step Working
1. Calculate the molar mass of Na2SO3
The molar mass of a compound is the sum of the atomic masses of its constituent elements multiplied by the number of atoms of each element in the molecular formula.
The molecular formula of sodium sulfite is Na2SO3.
| Element | Atomic Mass (g/mol) | Number of atoms | Mass contribution (g/mol) |
|---|---|---|---|
| Na | 22.99 | 2 | 2 x 22.99 = 45.98 |
| S | 32.07 | 1 | 1 x 32.07 = 32.07 |
| O | 16.00 | 3 | 3 x 16.00 = 48.00 |
| Total molar mass | 126.05 g/mol |
The provided molar mass of Na2SO3 is 126.05 g/mol, which matches our calculation.
2. Use the molar mass to convert mass of Na2SO3 to moles
We are given the mass of Na2SO3 is 99.6 mg. We need to convert this to moles using the molar mass calculated above.
Using the formula:
Moles = Mass (g) / Molar mass (g/mol)
Moles of Na2SO3 = 99.6 mg / 126.05 g/mol
= 0.0007892 mol
3. Determine moles of sodium ions using the chemical formula
From the balanced chemical formula, we see that 1 mole of Na2SO3 contains 2 moles of Na+.
Using stoichiometry:
If 0.0007892 mol Na2SO3 contains 2 mol Na+
Then 1 mol Na2SO3 contains 2 mol Na+
So 0.0007892 mol Na2SO3 contains
= (0.0007892 mol Na2SO3) x (2 mol Na+ / 1 mol Na2SO3)
= 0.0015784 mol Na+ ions
4. Convert moles of Na+ to number of ions
To convert moles to a number of particles, we use Avogadro’s number which is 6.022 x 10^23 particles/mol.
Number of Na+ ions = Moles of Na+ x (6.022 x 10^23 ions/mol)
= 0.0015784 mol x (6.022 x 10^23 ions/mol)
= 9.51 x 10^20 Na+ ions
Therefore, the number of sodium ions contained in 99.6 mg of Na2SO3 is 9.51 x 10^20
Conclusion
By using the provided molar mass of Na2SO3, we calculated the number of moles present in 99.6 mg of the compound. We then used the chemical formula to relate the number of moles of Na2SO3 to the number of moles of sodium ions. Finally, by applying Avogadro’s number, we determined the number of sodium ions present is 9.51 x 10^20. This step-by-step calculation demonstrates the use of stoichiometric relationships and conversion factors to derive the number of ions from a given mass of a compound. The concepts of molar mass, chemical formulas, and Avogadro’s number are applied. For chemical calculations, understanding these fundamental concepts is essential.
Summary
– The molar mass of Na2SO3 is 126.05 g/mol
– 99.6 mg Na2SO3 corresponds to 0.0007892 mol Na2SO3
– Based on the chemical formula, 0.0007892 mol Na2SO3 contains 0.0015784 mol Na+
– Using Avogadro’s number, 0.0015784 mol Na+ contains 9.51 x 10^20 ions
– Therefore, 99.6 mg Na2SO3 contains 9.51 x 10^20 sodium ions.
I hope this detailed explanation and step-by-step working has clearly shown how to calculate the number of sodium ions in a given mass of sodium sulfite. Let me know if you need any clarification or have additional questions!
Frequently Asked Questions
Why do we need to calculate the molar mass first?
Calculating the molar mass is the first step because it provides the molecular weight of Na2SO3 in grams/mole. This conversion factor is needed to convert the given mass in mg to moles using the relationship:
Moles = Mass (g) / Molar mass (g/mol)
Without knowing the molar mass, we cannot determine the number of moles.
Where does the chemical formula Na2SO3 come from?
The chemical formula of a compound gives the number and types of elements present in one molecular unit. Sodium sulfite is an ionic compound made up of two sodium ions (Na+) and one sulfite ion (SO32-).
The lowest whole number ratio of ions is two sodium ions for every one sulfite ion. This gives the formula Na2SO3. The formula comes from experimental determination of the compound’s composition.
Why is Avogadro’s number used?
Avogadro’s number (6.022 x 10^23 particles/mol) is used to convert moles, a unit of amount of substance, into a number of individual particles.
Since ions like Na+ and molecules are discrete particles, counting them requires relating moles to the number of entities present using Avogadro’s constant.
This conversion from moles to number of particles allows us to determine the actual number of sodium ions present in the given sample mass.
What if we were given mass in a different unit like grams?
The same process would apply if the mass were given in grams instead of milligrams. For example, if the question stated 0.0996 g Na2SO3 instead of 99.6 mg Na2SO3, the calculation steps would remain the same.
The molar mass would be used to convert 0.0996 g to moles first. Then the mole ratio from the formula and Avogadro’s number would give the number of sodium ions present.
The key is using the provided mass and molar mass to get moles, regardless of the mass units. The rest of the process remains unchanged.
Applications
Some examples of where a calculation like this may be useful include:
– Determining the amount of sodium present in food products as a nutrient or food additive.
– Quality control and purity analysis of chemicals like sodium sulfite.
– Understanding stoichiometric ratios in chemical reactions involving ionic compounds.
– Converting between mass composition and number of entities for pharmacological compounds.
– Balancing equations and calculating product yields for chemical processes.
– Estimating dose amounts in medical treatments involving ionic salts.
So in summary, this type of calculation has many real-world applications across nutrition, chemistry, medicine, and industry. The same principles can be applied to calculate ions or molecules present in different compounds.
