Find the Density at STP DyLiSrCuO

$D y L i S r C u O$

Step 1

To find the mass of $1$ mole of $D y L i S r C u O$ look up the atomic mass of each element and multiply it by the number of atoms contained in each element in the molecule.

mass of $1$ mole of $D y L i S r C u O$ $=$ $(mass of Dy)$ + $(mass of Li)$ + $(mass of Sr)$ + $(mass of Cu)$ + $(mass of O)$

Step 2

Fill in the atomic masses from the periodic table.

mass of $1$ mole of $D y L i S r C u O$ $=$ $(162.5)$ + $(6.94000005)$ + $(87.62000274)$ + $(63.54999923)$ + $(16)$

Step 3

Simplify the result.

Remove parentheses.

mass of $1$ mole of $D y L i S r C u O = 162.5 + 6.94000005 + 87.62000274 + 63.54999923 + 16 g$

Add $162.5$ and $6.94000005$ .

mass of $1$ mole of $D y L i S r C u O = 169.44000005 + 87.62000274 + 63.54999923 + 16 g$

Add $169.44000005$ and $87.62000274$ .

mass of $1$ mole of $D y L i S r C u O = 257.0600028 + 63.54999923 + 16 g$

Add $257.0600028$ and $63.54999923$ .

mass of $1$ mole of $D y L i S r C u O = 320.61000204 + 16 g$

Add $320.61000204$ and $16$ .

mass of $1$ mole of $D y L i S r C u O = 336.61000204 g$

Step 4

One mole of any gas occupies $22.4 L$ at STP (standard temperature and pressure).

$V = 22.4 L$

Step 5

Density is mass over volume.

$\frac{336.61000204 g}{22.4 L}$

Step 6

Factor $336.61000204$ out of $336.61000204 g$ .

$\frac{336.61000204 (g)}{22.4 L}$

Step 7

Factor $22.4$ out of $22.4 L$ .

$\frac{336.61000204 (g)}{22.4 (L)}$

Step 8

Separate fractions.

$\frac{336.61000204}{22.4} \cdot \frac{g}{L}$

Step 9

Divide $336.61000204$ by $22.4$ .

$15.02723223 \frac{g}{L}$

Step 10

Combine $15.02723223$ and $\frac{g}{L}$ .

$\frac{15.02723223 g}{L}$

Find the Density at STP DyLiSrCuO

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