How many moles of \(\displaystyle H\) are in \(\displaystyle 40,000 \,mg\,\) of \(\displaystyle H_{2}\,O\) - to three significant figures?

For \(\displaystyle H_{2}\,O\), finding via the periodic table (atomic weights):

H - \(\displaystyle 1.00794 (2) \,g\,\, H = 2.01588 \,g\,\, H\)

O - \(\displaystyle 15.9994 \,g\,\, O \)

\(\displaystyle 2.01588\,g\, \,H + 15.9994\,g\,\, O = 18.01528\,g\,\, H_{2}\,O \)

\(\displaystyle 40,000\,mg\, H_{2}\,O\,(\dfrac{ 0.001\,g\, H_{2}\,O}{ 1\,mg\,\, H_{2}\,O}))(\dfrac{1\,mol\,H_{2}\,O}{18.01528\,g\,\,H_{2}\,O }(\dfrac{2\,mol\,H}{1\,mol\,H_{2}\,O}) = 4,44 \,mol\,H \)

How many moles of \(\displaystyle Cu\) are in \(\displaystyle 100,000\,mg\) of \(\displaystyle CuO \) - to three significant figures?

__Find the molar mass (formula weight) of water.__For \(\displaystyle H_{2}\,O\), finding via the periodic table (atomic weights):

H - \(\displaystyle 1.00794 (2) \,g\,\, H = 2.01588 \,g\,\, H\)

O - \(\displaystyle 15.9994 \,g\,\, O \)

\(\displaystyle 2.01588\,g\, \,H + 15.9994\,g\,\, O = 18.01528\,g\,\, H_{2}\,O \)

__Final Calculation__\(\displaystyle 40,000\,mg\, H_{2}\,O\,(\dfrac{ 0.001\,g\, H_{2}\,O}{ 1\,mg\,\, H_{2}\,O}))(\dfrac{1\,mol\,H_{2}\,O}{18.01528\,g\,\,H_{2}\,O }(\dfrac{2\,mol\,H}{1\,mol\,H_{2}\,O}) = 4,44 \,mol\,H \)

__Your Turn__How many moles of \(\displaystyle Cu\) are in \(\displaystyle 100,000\,mg\) of \(\displaystyle CuO \) - to three significant figures?

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