Saturday, July 6, 2019

MOLE : Q.A.E 2

Q1: How many oxygen atoms are in 120.23 g Fe3(PO4)2?



Answer:

120.23g Fe3(PO4)2 contains 1.6203 x 1024 atoms O.

Explanation:

There are eight moles of oxygen atoms in each mole of Fe3(PO4)2. To answer this question, you must convert the given mass of Fe3(PO4)2 to moles of Fe3(PO4)2. Then multiply the moles of Fe3(PO4)2 times (8mol O)/(1 mol Fe3(PO4)2) to determine the number of moles of oxygen. Then multiply the moles of oxygen times 6.022×1023 atoms O.
Given/Known:
mass Fe3(PO4)2=120.23 g
molar mass Fe3(PO4)2=357.478g/mol
1 mole Fe3(PO4)2 contains 8 moles oxygen atoms
1 mole O atoms = 6.022×1023 atoms
Unknown:
number of oxygen atoms in 120.23g Fe3(PO4)2
Solution:
1. Convert given mass of Fe3(PO4)2 to moles.
120.23g Fe3(PO4)2×1mol Fe3(PO4)2357.478g Fe3(PO4)2=0.336328 mol Fe3(PO4)2
2. Convert moles Fe3(PO4)2 to moles O.
0.336328mol Fe3(PO4)2×8mol O1mol Fe3(PO4)2=2.690624 mol O
3. Convert moles O to atoms O.
2.690624mol O×6.022×1023atoms O1mol O=1.6203×1024atoms O (answer rounded to five significant figures)

SOURCE : https://socratic.org/questions/5473b86c581e2a14bf83682d



Q2: How many atoms of carbon are in 0.0235 gof cocaine hydrochloride, C17H22ClNO4?

Answer:

There are 7.08×1020.atoms C in 0.0235 g C17H22ClNO4.

Explanation:

The molecular formula for cocaine hydrochloride, C17H22ClNO4, tells us that one mole of C17H22ClNO4 contains 17 mol C.
0.0235 g C17H22ClNO4 represents a fraction of a mole.
In order to determine the mol C17H22ClNO34 in 0.0235 g C17H22ClNO4, divide its given mass by its molar mass by multiplying by its inverse.
The molar mass of C17H22ClNO4 is 339.816 g/mol.
https://www.ncbi.nlm.nih.gov/pccompound?term=%22COCAINE+HYDROCHLORIDE%22
Moles of Cocaine Hydrochloride
Divide its given mass by its molar mass by multiplying by its inverse.
0.0235g C17H22ClNO4×1mol C17H22ClNO4339.816g C17H22ClNO4=0.00006916 mol C17H22ClNO4
Atoms of Carbon
Multiply mol C17H22ClNO4 by 17 mols C divided by 1 mol C17H22ClNO4, then multiply by 6.022×1023 atoms/mol.
0.000069161mol C17H22ClNO4×17mol C1mol C17H22ClNO4=0.00117572 mol C
0.00117572mol C×6.022×1023atoms C1mol C
=7.08×1020atoms C, rounded to three significant figures

SOURCE :https://socratic.org/questions/592c925811ef6b2834ec19cb

No comments:

Post a Comment