To determine how many moles are in 2.3 grams of phosphorus, we begin by understanding the fundamental relationship between mass and moles in chemistry. This relationship is defined by the concept of molar mass, which is the mass of one mole of a substance. Molar mass is calculated by summing the atomic masses of all the atoms in a molecule, and it is typically expressed in grams per mole (g/mol).
Step 1: Identify the Molar Mass of Phosphorus
Phosphorus is a chemical element with the symbol P and atomic number 15. Still, its atomic mass, as listed on the periodic table, is approximately 30. 97 g/mol. Since phosphorus typically exists as individual atoms in its elemental form (not as molecules like P₄), we use the atomic mass directly for our calculation And it works..
Step 2: Use the Molar Mass to Convert Grams to Moles
The formula to convert grams to moles is:
$ \text{moles} = \frac{\text{mass (in grams)}}{\text{molar mass (in g/mol)}} $
Given:
- Mass of phosphorus = 2.3 grams
- Molar mass of phosphorus = 30.97 g/mol
Substitute into the formula:
$ \text{moles} = \frac{2.3}{30.97} \approx 0.0743 $
Step 3: Consider Significant Figures
The mass of phosphorus (2.Worth adding: 3 grams) has two significant figures, while the molar mass (30. That said, 97 g/mol) has four significant figures. In calculations, the result should be reported with the same number of significant figures as the least precise measurement That's the whole idea..
$ 0.0743 \approx 0.074 $
Final Answer
$ \boxed{0.074} $
Thus, 2.3 grams of phosphorus contains approximately 0.074 moles.
Step 4: Addressing Molecular Form and Allotropes
While the previous calculation assumes phosphorus exists as individual atoms (P), it is critical to recognize that elemental phosphorus commonly occurs in molecular forms, particularly as P₄ (tetrahedral molecules in white phosphorus) or other allotropes like red phosphorus. If the phosphorus sample in question is in the P₄ form, its molar mass must be adjusted accordingly. The molar mass of P₄ is:
$ \text{Molar mass of } \text{P}_4 = 4 \
×30.97 g/mol ≈ 123.Day to day, 3}{123. So 0186 , \text{mol} , (\text{or } 0. But 88} \approx 0. If the given mass of 2.And 3 grams refers to P₄, the calculation would instead yield:
$ \text{moles of P}_4 = \frac{2. Also, 88 g/mol. 019 , \text{mol with two significant figures}) Easy to understand, harder to ignore..
Still, the original problem does not specify the allotrope, and the periodic table value of 30.97 g/mol inherently refers to atomic phosphorus (P). Thus, the initial calculation remains valid unless explicitly stated otherwise Not complicated — just consistent..
Step 5: Importance of Context in Chemical Calculations
This example underscores the importance of clarifying the form of a substance in stoichiometric problems. While atomic phosphorus (P) is the default assumption when molar mass is derived from the periodic table, molecular forms like P₄ or P₂ (in certain compounds) require adjustments to the molar mass. Misinterpreting the allotrope could lead to significant errors, emphasizing the need for precise problem statements in real-world applications The details matter here..
Conclusion
In the absence of explicit information about phosphorus’ allotrope, the calculation assumes elemental phosphorus exists as individual atoms (P). Using the molar mass of 30.97 g/mol, 2.3 grams of phosphorus equals 0.074 moles. This result highlights the critical role of molar mass in converting between mass and moles and the necessity of considering significant figures for precision. Always verify whether a substance is presented in its atomic, molecular, or compound form to ensure accurate stoichiometric calculations It's one of those things that adds up..
Final Answer:
\boxed{0.074}
